Average Velocity
1
A particle moves from A to B, covering 10 m in 5 seconds towards the east. What are its distance, displacement, average speed and average velocity?
Practice
A
Distance = 10 m, Displacement = 10 m east, Average speed = 2 m/s, Average velocity = 2 m/s east
B
Distance = 10 m, Displacement = 0 m, Average speed = 2 m/s, Average velocity = 0 m/s
C
Distance = 5 m, Displacement = 10 m east, Average speed = 5 m/s, Average velocity = 2 m/s east
D
Distance = 10 m, Displacement = 5 m east, Average speed = 2 m/s, Average velocity = 1 m/s east
β Show Answer
β Correct Answer:
A
(Distance = 10 m, Displacement = 10 m east, Average speed = 2 m/s, Average velocity = 2 m/s east)
π‘ Explanation
Given:
Distance travelled = 10 m
Time = 5 s
Direction = East
Since the particle moves straight from A to B without changing direction:
Displacement = 10 m east
Average speed = Total distance / Total time
= 10 / 5
= 2 m/s
Average velocity = Total displacement / Total time
= 10 / 5
= 2 m/s east
Therefore:
Distance = 10 m
Displacement = 10 m east
Average speed = 2 m/s
Average velocity = 2 m/s east.
Distance travelled = 10 m
Time = 5 s
Direction = East
Since the particle moves straight from A to B without changing direction:
Displacement = 10 m east
Average speed = Total distance / Total time
= 10 / 5
= 2 m/s
Average velocity = Total displacement / Total time
= 10 / 5
= 2 m/s east
Therefore:
Distance = 10 m
Displacement = 10 m east
Average speed = 2 m/s
Average velocity = 2 m/s east.
2
A particle moves from A to B, covering 10 m in 5 seconds towards the east, and then moves 10 m towards the west in 10 seconds. Find the total distance, displacement, average speed and average velocity.
Practice
A
Distance = 20 m, Displacement = 0 m, Average speed = 4/3 m/s, Average velocity = 0 m/s
B
Distance = 20 m, Displacement = 20 m east, Average speed = 4/3 m/s, Average velocity = 4/3 m/s east
C
Distance = 0 m, Displacement = 20 m east, Average speed = 0 m/s, Average velocity = 0 m/s
D
Distance = 20 m, Displacement = 10 m east, Average speed = 2 m/s, Average velocity = 2/3 m/s east
β Show Answer
β Correct Answer:
A
(Distance = 20 m, Displacement = 0 m, Average speed = 4/3 m/s, Average velocity = 0 m/s)
π‘ Explanation
Given:
First motion: 10 m east in 5 s
Second motion: 10 m west in 10 s
Total distance = 10 + 10
= 20 m
The particle returns to its starting point.
Therefore, displacement = 0 m
Total time = 5 + 10
= 15 s
Average speed = Total distance / Total time
= 20 / 15
= 4/3 m/s
Average velocity = Total displacement / Total time
= 0 / 15
= 0 m/s
Therefore:
Distance = 20 m
Displacement = 0 m
Average speed = 4/3 m/s
Average velocity = 0 m/s.
First motion: 10 m east in 5 s
Second motion: 10 m west in 10 s
Total distance = 10 + 10
= 20 m
The particle returns to its starting point.
Therefore, displacement = 0 m
Total time = 5 + 10
= 15 s
Average speed = Total distance / Total time
= 20 / 15
= 4/3 m/s
Average velocity = Total displacement / Total time
= 0 / 15
= 0 m/s
Therefore:
Distance = 20 m
Displacement = 0 m
Average speed = 4/3 m/s
Average velocity = 0 m/s.
3
A particle moves from A to B, covering 20 m in 5 seconds towards the east, and then moves 30 m towards the west in 60 seconds. Find the total distance, displacement, average speed and average velocity.
Practice
A
Distance = 50 m, Displacement = 10 m west, Average speed = 5/6 m/s, Average velocity = 1/13 m/s west
B
Distance = 50 m, Displacement = 10 m east, Average speed = 5/6 m/s, Average velocity = 1/13 m/s east
C
Distance = 10 m, Displacement = 50 m west, Average speed = 1/13 m/s, Average velocity = 5/6 m/s west
D
Distance = 50 m, Displacement = 50 m west, Average speed = 5/6 m/s, Average velocity = 5/6 m/s west
β Show Answer
β Correct Answer:
A
(Distance = 50 m, Displacement = 10 m west, Average speed = 5/6 m/s, Average velocity = 1/13 m/s west)
π‘ Explanation
Given:
First motion = 20 m east in 5 s
Second motion = 30 m west in 60 s
Take east as positive and west as negative.
Total distance = 20 + 30
= 50 m
Displacement = +20 - 30
= -10 m
Therefore, displacement = 10 m west.
Total time = 5 + 60
= 65 s
Average speed = Total distance / Total time
= 50 / 65
= 10/13 m/s
β 0.77 m/s
Average velocity = Total displacement / Total time
= -10 / 65
= -2/13 m/s
β -0.154 m/s
The negative sign indicates west.
Therefore:
Distance = 50 m
Displacement = 10 m west
Average speed = 10/13 m/s
Average velocity = 2/13 m/s west.
First motion = 20 m east in 5 s
Second motion = 30 m west in 60 s
Take east as positive and west as negative.
Total distance = 20 + 30
= 50 m
Displacement = +20 - 30
= -10 m
Therefore, displacement = 10 m west.
Total time = 5 + 60
= 65 s
Average speed = Total distance / Total time
= 50 / 65
= 10/13 m/s
β 0.77 m/s
Average velocity = Total displacement / Total time
= -10 / 65
= -2/13 m/s
β -0.154 m/s
The negative sign indicates west.
Therefore:
Distance = 50 m
Displacement = 10 m west
Average speed = 10/13 m/s
Average velocity = 2/13 m/s west.
4
A particle moves 10 m in the +x direction in 5 seconds, and then moves 20 m in the +y direction in 20 seconds. What are the total distance, total displacement, average speed and average velocity?
Practice
A
30 m, 10β5 m, 1.2 m/s, (2β5)/5 m/s at 63.4Β° above +x-axis
B
30 m, 30 m, 1.2 m/s, 1.2 m/s
C
30 m, 20 m, 1.2 m/s, 0.8 m/s
D
10β5 m, 30 m, 0.8 m/s, (2β5)/5 m/s
β Show Answer
β Correct Answer:
A
(30 m, 10β5 m, 1.2 m/s, (2β5)/5 m/s at 63.4Β° above +x-axis)
π‘ Explanation
Given:
First motion = 10 m in +x direction in 5 s
Second motion = 20 m in +y direction in 20 s
Total distance = 10 + 20
= 30 m
Since the two displacements are perpendicular:
Total displacement = β(10Β² + 20Β²)
= β(100 + 400)
= β500
= 10β5 m
β 22.36 m
Total time = 5 + 20
= 25 s
Average speed = Total distance / Total time
= 30 / 25
= 1.2 m/s
Average velocity = Total displacement / Total time
= 10β5 / 25
= 2β5 / 5 m/s
β 0.894 m/s
Direction of average velocity:
tan ΞΈ = 20 / 10 = 2
ΞΈ = tanβ»ΒΉ(2) β 63.4Β°
Therefore:
Total distance = 30 m
Total displacement = 10β5 m β 22.36 m
Average speed = 1.2 m/s
Average velocity = (2β5)/5 m/s β 0.894 m/s at 63.4Β° above the +x-axis.
First motion = 10 m in +x direction in 5 s
Second motion = 20 m in +y direction in 20 s
Total distance = 10 + 20
= 30 m
Since the two displacements are perpendicular:
Total displacement = β(10Β² + 20Β²)
= β(100 + 400)
= β500
= 10β5 m
β 22.36 m
Total time = 5 + 20
= 25 s
Average speed = Total distance / Total time
= 30 / 25
= 1.2 m/s
Average velocity = Total displacement / Total time
= 10β5 / 25
= 2β5 / 5 m/s
β 0.894 m/s
Direction of average velocity:
tan ΞΈ = 20 / 10 = 2
ΞΈ = tanβ»ΒΉ(2) β 63.4Β°
Therefore:
Total distance = 30 m
Total displacement = 10β5 m β 22.36 m
Average speed = 1.2 m/s
Average velocity = (2β5)/5 m/s β 0.894 m/s at 63.4Β° above the +x-axis.
5
A particle moves in a circle of radius 10 m and completes 2 complete revolutions, returning to its initial position. Each revolution takes 10 seconds. What are the total distance, total displacement, average speed and average velocity?
Practice
A
Distance = 40Ο m, Displacement = 0 m, Average speed = 4Ο m/s, Average velocity = 0 m/s
B
Distance = 20Ο m, Displacement = 0 m, Average speed = 2Ο m/s, Average velocity = 0 m/s
C
Distance = 40Ο m, Displacement = 20 m, Average speed = 4Ο m/s, Average velocity = 2 m/s
D
Distance = 20Ο m, Displacement = 20 m, Average speed = 2Ο m/s, Average velocity = 2 m/s
β Show Answer
β Correct Answer:
A
(Distance = 40Ο m, Displacement = 0 m, Average speed = 4Ο m/s, Average velocity = 0 m/s)
π‘ Explanation
Given:
Radius of circle = 10 m
Number of revolutions = 2
Time for each revolution = 10 s
Circumference of the circle = 2Οr
= 2Ο Γ 10
= 20Ο m
Total distance = Number of revolutions Γ Circumference
= 2 Γ 20Ο
= 40Ο m
After 2 complete revolutions, the particle returns to its initial position.
Therefore, total displacement = 0 m
Total time = 2 Γ 10
= 20 s
Average speed = Total distance / Total time
= 40Ο / 20
= 2Ο m/s
Average velocity = Total displacement / Total time
= 0 / 20
= 0 m/s
Therefore:
Total distance = 40Ο m
Total displacement = 0 m
Average speed = 2Ο m/s
Average velocity = 0 m/s.
Radius of circle = 10 m
Number of revolutions = 2
Time for each revolution = 10 s
Circumference of the circle = 2Οr
= 2Ο Γ 10
= 20Ο m
Total distance = Number of revolutions Γ Circumference
= 2 Γ 20Ο
= 40Ο m
After 2 complete revolutions, the particle returns to its initial position.
Therefore, total displacement = 0 m
Total time = 2 Γ 10
= 20 s
Average speed = Total distance / Total time
= 40Ο / 20
= 2Ο m/s
Average velocity = Total displacement / Total time
= 0 / 20
= 0 m/s
Therefore:
Total distance = 40Ο m
Total displacement = 0 m
Average speed = 2Ο m/s
Average velocity = 0 m/s.
6
A particle moves through one-fourth of a circle of radius 10 m in 10 seconds. Find the total distance, total displacement, average speed and average velocity.
Practice
A
Distance = 5Ο m, Displacement = 10β2 m, Average speed = Ο/2 m/s, Average velocity = β2 m/s
B
Distance = 10Ο m, Displacement = 20 m, Average speed = Ο m/s, Average velocity = 2 m/s
C
Distance = 5Ο m, Displacement = 10 m, Average speed = Ο/2 m/s, Average velocity = 1 m/s
D
Distance = 20Ο m, Displacement = 10β2 m, Average speed = 2Ο m/s, Average velocity = β2/2 m/s
β Show Answer
β Correct Answer:
A
(Distance = 5Ο m, Displacement = 10β2 m, Average speed = Ο/2 m/s, Average velocity = β2 m/s)
π‘ Explanation
Given:
Radius of circle = 10 m
Fraction of circle travelled = 1/4
Time = 10 s
Circumference of the circle = 2Οr
= 2Ο Γ 10
= 20Ο m
Distance travelled in 1/4 circle:
Distance = (1/4) Γ 20Ο
= 5Ο m
For 1/4 of a circle, the angle at the centre is 90Β°.
The displacement is the chord joining the initial and final positions.
Displacement = β(10Β² + 10Β²)
= β200
= 10β2 m
Average speed = Total distance / Total time
= 5Ο / 10
= Ο/2 m/s
Average velocity = Total displacement / Total time
= 10β2 / 10
= β2 m/s
Therefore:
Distance = 5Ο m
Displacement = 10β2 m
Average speed = Ο/2 m/s
Average velocity = β2 m/s.
Radius of circle = 10 m
Fraction of circle travelled = 1/4
Time = 10 s
Circumference of the circle = 2Οr
= 2Ο Γ 10
= 20Ο m
Distance travelled in 1/4 circle:
Distance = (1/4) Γ 20Ο
= 5Ο m
For 1/4 of a circle, the angle at the centre is 90Β°.
The displacement is the chord joining the initial and final positions.
Displacement = β(10Β² + 10Β²)
= β200
= 10β2 m
Average speed = Total distance / Total time
= 5Ο / 10
= Ο/2 m/s
Average velocity = Total displacement / Total time
= 10β2 / 10
= β2 m/s
Therefore:
Distance = 5Ο m
Displacement = 10β2 m
Average speed = Ο/2 m/s
Average velocity = β2 m/s.
7
A particle moves through one-half of a circle of radius 10 m in 10 seconds. Find the total distance, total displacement, average speed and average velocity.
Practice
A
Distance = 10Ο m, Displacement = 20 m, Average speed = Ο m/s, Average velocity = 2 m/s
B
Distance = 20Ο m, Displacement = 10 m, Average speed = 2Ο m/s, Average velocity = 1 m/s
C
Distance = 10Ο m, Displacement = 0 m, Average speed = Ο m/s, Average velocity = 0 m/s
D
Distance = 20 m, Displacement = 10Ο m, Average speed = 2 m/s, Average velocity = Ο m/s
β Show Answer
β Correct Answer:
A
(Distance = 10Ο m, Displacement = 20 m, Average speed = Ο m/s, Average velocity = 2 m/s)
π‘ Explanation
Given:
Radius = 10 m
Time = 10 s
Fraction of circle travelled = 1/2
Circumference of circle = 2Οr
= 2Ο Γ 10
= 20Ο m
Total distance = (1/2) Γ circumference
= (1/2) Γ 20Ο
= 10Ο m
For half a circle, the initial and final points are opposite ends of the diameter.
Therefore, displacement = diameter
= 2r
= 2 Γ 10
= 20 m
Average speed = Total distance / Total time
= 10Ο / 10
= Ο m/s
Average velocity = Total displacement / Total time
= 20 / 10
= 2 m/s
Therefore:
Total distance = 10Ο m
Total displacement = 20 m
Average speed = Ο m/s
Average velocity = 2 m/s.
Radius = 10 m
Time = 10 s
Fraction of circle travelled = 1/2
Circumference of circle = 2Οr
= 2Ο Γ 10
= 20Ο m
Total distance = (1/2) Γ circumference
= (1/2) Γ 20Ο
= 10Ο m
For half a circle, the initial and final points are opposite ends of the diameter.
Therefore, displacement = diameter
= 2r
= 2 Γ 10
= 20 m
Average speed = Total distance / Total time
= 10Ο / 10
= Ο m/s
Average velocity = Total displacement / Total time
= 20 / 10
= 2 m/s
Therefore:
Total distance = 10Ο m
Total displacement = 20 m
Average speed = Ο m/s
Average velocity = 2 m/s.
8
A particle moves through 3/4th of a circle of radius 10 m in 10 seconds. Find the total distance, total displacement, average speed and average velocity.
Practice
A
Distance = 15Ο m, Displacement = 10β2 m, Average speed = 3Ο/2 m/s, Average velocity = β2 m/s
B
Distance = 15Ο m, Displacement = 20 m, Average speed = 3Ο/2 m/s, Average velocity = 2 m/s
C
Distance = 20Ο m, Displacement = 10β2 m, Average speed = 2Ο m/s, Average velocity = β2 m/s
D
Distance = 10Ο m, Displacement = 15 m, Average speed = Ο m/s, Average velocity = 1.5 m/s
β Show Answer
β Correct Answer:
A
(Distance = 15Ο m, Displacement = 10β2 m, Average speed = 3Ο/2 m/s, Average velocity = β2 m/s)
π‘ Explanation
Given:
Radius = 10 m
Fraction of circle travelled = 3/4
Time = 10 s
Circumference of circle = 2Οr
= 2Ο Γ 10
= 20Ο m
Total distance = (3/4) Γ circumference
= (3/4) Γ 20Ο
= 15Ο m
For 3/4 of a circle, the angle between the initial and final radius is 90Β°.
Therefore, the displacement is the chord joining the initial and final positions.
Displacement = β(10Β² + 10Β²)
= β200
= 10β2 m
Total time = 10 s
Average speed = Total distance / Total time
= 15Ο / 10
= 3Ο/2 m/s
Average velocity = Total displacement / Total time
= 10β2 / 10
= β2 m/s
Therefore:
Total distance = 15Ο m
Total displacement = 10β2 m
Average speed = 3Ο/2 m/s
Average velocity = β2 m/s.
Radius = 10 m
Fraction of circle travelled = 3/4
Time = 10 s
Circumference of circle = 2Οr
= 2Ο Γ 10
= 20Ο m
Total distance = (3/4) Γ circumference
= (3/4) Γ 20Ο
= 15Ο m
For 3/4 of a circle, the angle between the initial and final radius is 90Β°.
Therefore, the displacement is the chord joining the initial and final positions.
Displacement = β(10Β² + 10Β²)
= β200
= 10β2 m
Total time = 10 s
Average speed = Total distance / Total time
= 15Ο / 10
= 3Ο/2 m/s
Average velocity = Total displacement / Total time
= 10β2 / 10
= β2 m/s
Therefore:
Total distance = 15Ο m
Total displacement = 10β2 m
Average speed = 3Ο/2 m/s
Average velocity = β2 m/s.
9
A particle moves through an arc of 135Β° of a circle of radius 10 m in 10 seconds. Find the total distance, total displacement, average speed and average velocity.
Practice
A
Distance = 15Ο/2 m, Displacement = 10β(2 + β2) m, Average speed = 3Ο/4 m/s, Average velocity = β(2 + β2) m/s
B
Distance = 15Ο m, Displacement = 20 m, Average speed = 3Ο/2 m/s, Average velocity = 2 m/s
C
Distance = 15Ο/2 m, Displacement = 10β2 m, Average speed = 3Ο/4 m/s, Average velocity = β2 m/s
D
Distance = 10Ο m, Displacement = 15 m, Average speed = Ο m/s, Average velocity = 1.5 m/s
β Show Answer
β Correct Answer:
A
(Distance = 15Ο/2 m, Displacement = 10β(2 + β2) m, Average speed = 3Ο/4 m/s, Average velocity = β(2 + β2) m/s)
π‘ Explanation
Given:
Radius = 10 m
Angle travelled = 135Β°
Time = 10 s
Total distance = Arc length
= (ΞΈ/360Β°) Γ 2Οr
= (135/360) Γ 2Ο Γ 10
= (3/8) Γ 20Ο
= 15Ο/2 m
For displacement, use the chord joining the initial and final positions.
Displacement = 2r sin(ΞΈ/2)
= 2 Γ 10 Γ sin(135Β°/2)
= 20 sin(67.5Β°)
= 10β(2 + β2) m
β 18.48 m
Average speed = Total distance / Total time
= (15Ο/2) / 10
= 3Ο/4 m/s
β 2.36 m/s
Average velocity = Total displacement / Total time
= [10β(2 + β2)] / 10
= β(2 + β2) m/s
β 1.85 m/s
Therefore:
Total distance = 15Ο/2 m
Total displacement = 10β(2 + β2) m β 18.48 m
Average speed = 3Ο/4 m/s β 2.36 m/s
Average velocity = β(2 + β2) m/s β 1.85 m/s.
Radius = 10 m
Angle travelled = 135Β°
Time = 10 s
Total distance = Arc length
= (ΞΈ/360Β°) Γ 2Οr
= (135/360) Γ 2Ο Γ 10
= (3/8) Γ 20Ο
= 15Ο/2 m
For displacement, use the chord joining the initial and final positions.
Displacement = 2r sin(ΞΈ/2)
= 2 Γ 10 Γ sin(135Β°/2)
= 20 sin(67.5Β°)
= 10β(2 + β2) m
β 18.48 m
Average speed = Total distance / Total time
= (15Ο/2) / 10
= 3Ο/4 m/s
β 2.36 m/s
Average velocity = Total displacement / Total time
= [10β(2 + β2)] / 10
= β(2 + β2) m/s
β 1.85 m/s
Therefore:
Total distance = 15Ο/2 m
Total displacement = 10β(2 + β2) m β 18.48 m
Average speed = 3Ο/4 m/s β 2.36 m/s
Average velocity = β(2 + β2) m/s β 1.85 m/s.
10
A particle moves from position x = 20 m to x = 80 m in 6 s. It then moves back to x = 50 m in the next 4 s. What are the total distance, total displacement, average speed and average velocity of the particle for the entire motion?
Practice
A
Distance = 90 m, Displacement = 30 m, Average speed = 9 m/s, Average velocity = 3 m/s
B
Distance = 30 m, Displacement = 90 m, Average speed = 3 m/s, Average velocity = 9 m/s
C
Distance = 90 m, Displacement = 30 m, Average speed = 3 m/s, Average velocity = 9 m/s
D
Distance = 60 m, Displacement = 30 m, Average speed = 6 m/s, Average velocity = 3 m/s
β Show Answer
β Correct Answer:
A
(Distance = 90 m, Displacement = 30 m, Average speed = 9 m/s, Average velocity = 3 m/s)
π‘ Explanation
Given:
Initial position = 20 m
First final position = 80 m
Final position = 50 m
First time = 6 s
Second time = 4 s
First part:
Distance = 80 - 20
= 60 m
Second part:
The particle moves back from 80 m to 50 m.
Distance = 80 - 50
= 30 m
Total distance = 60 + 30
= 90 m
Total displacement = Final position - Initial position
= 50 - 20
= 30 m
Total time = 6 + 4
= 10 s
Average speed = Total distance / Total time
= 90 / 10
= 9 m/s
Average velocity = Total displacement / Total time
= 30 / 10
= 3 m/s
Therefore:
Total distance = 90 m
Total displacement = 30 m
Average speed = 9 m/s
Average velocity = 3 m/s.
Initial position = 20 m
First final position = 80 m
Final position = 50 m
First time = 6 s
Second time = 4 s
First part:
Distance = 80 - 20
= 60 m
Second part:
The particle moves back from 80 m to 50 m.
Distance = 80 - 50
= 30 m
Total distance = 60 + 30
= 90 m
Total displacement = Final position - Initial position
= 50 - 20
= 30 m
Total time = 6 + 4
= 10 s
Average speed = Total distance / Total time
= 90 / 10
= 9 m/s
Average velocity = Total displacement / Total time
= 30 / 10
= 3 m/s
Therefore:
Total distance = 90 m
Total displacement = 30 m
Average speed = 9 m/s
Average velocity = 3 m/s.
11
A particle travels half of the total distance with a speed of 20 m/s and the remaining half of the distance with a speed of 40 m/s. Assuming both halves are along the same straight line in the same direction, find the average speed and average velocity.
Practice
A
Average speed = 80/3 m/s, Average velocity = 80/3 m/s
B
Average speed = 30 m/s, Average velocity = 30 m/s
C
Average speed = 40/3 m/s, Average velocity = 20 m/s
D
Average speed = 25 m/s, Average velocity = 25 m/s
β Show Answer
β Correct Answer:
A
(Average speed = 80/3 m/s, Average velocity = 80/3 m/s)
π‘ Explanation
Let the total distance be 2d.
First half:
Distance = d
Speed = 20 m/s
Time = d/20
Second half:
Distance = d
Speed = 40 m/s
Time = d/40
Total distance = 2d
Total time = d/20 + d/40
= 3d/40
Average speed = Total distance / Total time
= 2d / (3d/40)
= 80/3 m/s
β 26.67 m/s
Since both halves are along the same straight line in the same direction:
Total displacement = Total distance = 2d
Average velocity = Total displacement / Total time
= 2d / (3d/40)
= 80/3 m/s
β 26.67 m/s
Therefore:
Average speed = 80/3 m/s β 26.67 m/s
Average velocity = 80/3 m/s β 26.67 m/s.
First half:
Distance = d
Speed = 20 m/s
Time = d/20
Second half:
Distance = d
Speed = 40 m/s
Time = d/40
Total distance = 2d
Total time = d/20 + d/40
= 3d/40
Average speed = Total distance / Total time
= 2d / (3d/40)
= 80/3 m/s
β 26.67 m/s
Since both halves are along the same straight line in the same direction:
Total displacement = Total distance = 2d
Average velocity = Total displacement / Total time
= 2d / (3d/40)
= 80/3 m/s
β 26.67 m/s
Therefore:
Average speed = 80/3 m/s β 26.67 m/s
Average velocity = 80/3 m/s β 26.67 m/s.
12
A particle travels half of the total distance with a speed of 20 m/s and the remaining half of the distance with a speed of 40 m/s. Assuming the two halves are along the same straight line but in opposite directions, find the average speed and average velocity.
Practice
A
Average speed = 80/3 m/s, Average velocity = 0 m/s
B
Average speed = 80/3 m/s, Average velocity = -40/3 m/s
C
Average speed = 30 m/s, Average velocity = 10 m/s
D
Average speed = 40/3 m/s, Average velocity = 20 m/s
β Show Answer
β Correct Answer:
A
(Average speed = 80/3 m/s, Average velocity = 0 m/s)
π‘ Explanation
Let the total distance be 2d.
First half:
Distance = d
Speed = 20 m/s
Time = d/20
Second half:
Distance = d
Speed = 40 m/s
Time = d/40
Total distance = 2d
Total time = d/20 + d/40
= 3d/40
Average speed = Total distance / Total time
= 2d / (3d/40)
= 80/3 m/s
β 26.67 m/s
Take the direction of the first half as positive.
First displacement = +d
Second displacement = -d
Total displacement = +d - d
= 0
Average velocity = Total displacement / Total time
= 0 / (3d/40)
= 0 m/s
Therefore:
Average speed = 80/3 m/s β 26.67 m/s
Average velocity = 0 m/s.
First half:
Distance = d
Speed = 20 m/s
Time = d/20
Second half:
Distance = d
Speed = 40 m/s
Time = d/40
Total distance = 2d
Total time = d/20 + d/40
= 3d/40
Average speed = Total distance / Total time
= 2d / (3d/40)
= 80/3 m/s
β 26.67 m/s
Take the direction of the first half as positive.
First displacement = +d
Second displacement = -d
Total displacement = +d - d
= 0
Average velocity = Total displacement / Total time
= 0 / (3d/40)
= 0 m/s
Therefore:
Average speed = 80/3 m/s β 26.67 m/s
Average velocity = 0 m/s.
13
A particle travels 1/3 of the total distance with a speed of 20 m/s and the remaining 2/3 of the distance with a speed of 40 m/s. Assuming both parts are along the same straight line in the same direction, find the average speed and average velocity.
Practice
A
Average speed = 30 m/s, Average velocity = 30 m/s
B
Average speed = 32 m/s, Average velocity = 32 m/s
C
Average speed = 35 m/s, Average velocity = 35 m/s
D
Average speed = 40/3 m/s, Average velocity = 40/3 m/s
β Show Answer
β Correct Answer:
B
(Average speed = 32 m/s, Average velocity = 32 m/s)
π‘ Explanation
Let the total distance be 3d.
First part:
Distance = d
Speed = 20 m/s
Time = d/20
Remaining part:
Distance = 2d
Speed = 40 m/s
Time = 2d/40
= d/20
Total distance = 3d
Total time = d/20 + d/20
= d/10
Average speed = Total distance / Total time
= 3d / (d/10)
= 30 m/s
Since both parts are along the same straight line in the same direction:
Total displacement = Total distance = 3d
Average velocity = Total displacement / Total time
= 3d / (d/10)
= 30 m/s
Therefore:
Average speed = 30 m/s
Average velocity = 30 m/s.
First part:
Distance = d
Speed = 20 m/s
Time = d/20
Remaining part:
Distance = 2d
Speed = 40 m/s
Time = 2d/40
= d/20
Total distance = 3d
Total time = d/20 + d/20
= d/10
Average speed = Total distance / Total time
= 3d / (d/10)
= 30 m/s
Since both parts are along the same straight line in the same direction:
Total displacement = Total distance = 3d
Average velocity = Total displacement / Total time
= 3d / (d/10)
= 30 m/s
Therefore:
Average speed = 30 m/s
Average velocity = 30 m/s.
14
A particle travels three equal distances with speeds of 20 m/s, 40 m/s and 60 m/s respectively. Assuming all three parts are along the same straight line in the same direction, find the average speed and average velocity.
Practice
A
Average speed = 180/7 m/s, Average velocity = 180/7 m/s
B
Average speed = 36 m/s, Average velocity = 36 m/s
C
Average speed = 40 m/s, Average velocity = 40 m/s
D
Average speed = 120/7 m/s, Average velocity = 120/7 m/s
β Show Answer
β Correct Answer:
A
(Average speed = 180/7 m/s, Average velocity = 180/7 m/s)
π‘ Explanation
Let each equal distance be d.
First part:
Distance = d, Speed = 20 m/s
Time = d/20
Second part:
Distance = d, Speed = 40 m/s
Time = d/40
Third part:
Distance = d, Speed = 60 m/s
Time = d/60
Total distance = 3d
Total time = d/20 + d/40 + d/60
= (6d + 3d + 2d)/120
= 11d/120
Average speed = Total distance / Total time
= 3d / (11d/120)
= 360/11 m/s
β 32.73 m/s
Since all three parts are along the same straight line in the same direction:
Total displacement = Total distance = 3d
Average velocity = Total displacement / Total time
= 3d / (11d/120)
= 360/11 m/s
β 32.73 m/s
Therefore:
Average speed = 360/11 m/s β 32.73 m/s
Average velocity = 360/11 m/s β 32.73 m/s.
First part:
Distance = d, Speed = 20 m/s
Time = d/20
Second part:
Distance = d, Speed = 40 m/s
Time = d/40
Third part:
Distance = d, Speed = 60 m/s
Time = d/60
Total distance = 3d
Total time = d/20 + d/40 + d/60
= (6d + 3d + 2d)/120
= 11d/120
Average speed = Total distance / Total time
= 3d / (11d/120)
= 360/11 m/s
β 32.73 m/s
Since all three parts are along the same straight line in the same direction:
Total displacement = Total distance = 3d
Average velocity = Total displacement / Total time
= 3d / (11d/120)
= 360/11 m/s
β 32.73 m/s
Therefore:
Average speed = 360/11 m/s β 32.73 m/s
Average velocity = 360/11 m/s β 32.73 m/s.
15
A particle travels 1/3rd of the total distance with a speed of 20 m/s, and the remaining distance is divided equally into two parts, which are travelled at speeds of 40 m/s and 60 m/s respectively. Assuming all three parts are along the same straight line in the same direction, find the average speed and average velocity.
Practice
A
Average speed = 360/11 m/s, Average velocity = 360/11 m/s
B
Average speed = 40 m/s, Average velocity = 40 m/s
C
Average speed = 30 m/s, Average velocity = 30 m/s
D
Average speed = 180/11 m/s, Average velocity = 180/11 m/s
β Show Answer
β Correct Answer:
A
(Average speed = 360/11 m/s, Average velocity = 360/11 m/s)
π‘ Explanation
Let the total distance be 3d.
First part:
Distance = d
Speed = 20 m/s
Time = d/20
Second part:
Distance = d
Speed = 40 m/s
Time = d/40
Third part:
Distance = d
Speed = 60 m/s
Time = d/60
Total distance = 3d
Total time = d/20 + d/40 + d/60
= (6d + 3d + 2d)/120
= 11d/120
Average speed = Total distance / Total time
= 3d / (11d/120)
= 360/11 m/s
β 32.73 m/s
Since all three parts are along the same straight line in the same direction:
Total displacement = Total distance = 3d
Average velocity = Total displacement / Total time
= 3d / (11d/120)
= 360/11 m/s
β 32.73 m/s
Therefore:
Average speed = 360/11 m/s β 32.73 m/s
Average velocity = 360/11 m/s β 32.73 m/s.
First part:
Distance = d
Speed = 20 m/s
Time = d/20
Second part:
Distance = d
Speed = 40 m/s
Time = d/40
Third part:
Distance = d
Speed = 60 m/s
Time = d/60
Total distance = 3d
Total time = d/20 + d/40 + d/60
= (6d + 3d + 2d)/120
= 11d/120
Average speed = Total distance / Total time
= 3d / (11d/120)
= 360/11 m/s
β 32.73 m/s
Since all three parts are along the same straight line in the same direction:
Total displacement = Total distance = 3d
Average velocity = Total displacement / Total time
= 3d / (11d/120)
= 360/11 m/s
β 32.73 m/s
Therefore:
Average speed = 360/11 m/s β 32.73 m/s
Average velocity = 360/11 m/s β 32.73 m/s.
16
A particle moves along a straight line. Its position changes from x = 10 m to x = 40 m in 5 s and then from x = 40 m to x = 10 m in 3 s. What is the average velocity during the first 8 s?
Practice
A
0 m/s
B
2.5 m/s
C
3.75 m/s
D
7.5 m/s
β Show Answer
β Correct Answer:
A
(0 m/s)
π‘ Explanation
Average velocity = Total displacement / Total time
Initial position = 10 m
Final position = 10 m
Total displacement = Final position β Initial position
= 10 β 10
= 0 m
Total time = 5 + 3
= 8 s
Average velocity = 0 / 8
= 0 m/s
Therefore, the correct answer is 0 m/s.
Important concept:
The particle returns to its initial position, so its net displacement is zero. Therefore, its average velocity is zero, even though it has travelled a non-zero distance.
Initial position = 10 m
Final position = 10 m
Total displacement = Final position β Initial position
= 10 β 10
= 0 m
Total time = 5 + 3
= 8 s
Average velocity = 0 / 8
= 0 m/s
Therefore, the correct answer is 0 m/s.
Important concept:
The particle returns to its initial position, so its net displacement is zero. Therefore, its average velocity is zero, even though it has travelled a non-zero distance.
17
A particle moves along the x-axis. Its position is x = 5 m at t = 2 s and x = 29 m at t = 6 s. What is the average velocity between t = 2 s and t = 6 s?
Practice
A
4 m/s
B
6 m/s
C
8 m/s
D
12 m/s
β Show Answer
β Correct Answer:
B
(6 m/s)
π‘ Explanation
Average velocity = Change in position / Change in time
Initial position:
xβ = 5 m at tβ = 2 s
Final position:
xβ = 29 m at tβ = 6 s
Change in position = xβ β xβ
= 29 β 5
= 24 m
Change in time = tβ β tβ
= 6 β 2
= 4 s
Average velocity = 24 / 4
= 6 m/s
Therefore, the correct answer is 6 m/s.
Initial position:
xβ = 5 m at tβ = 2 s
Final position:
xβ = 29 m at tβ = 6 s
Change in position = xβ β xβ
= 29 β 5
= 24 m
Change in time = tβ β tβ
= 6 β 2
= 4 s
Average velocity = 24 / 4
= 6 m/s
Therefore, the correct answer is 6 m/s.
18
A particle moves from x = 0 m to x = 40 m in 4 s. It then remains at rest for 2 s and finally moves to x = 70 m in 3 s. What is the average velocity for the entire motion?
Practice
A
5 m/s
B
7 m/s
C
7.78 m/s
D
10 m/s
β Show Answer
β Correct Answer:
C
(7.78 m/s)
π‘ Explanation
Average velocity = Total displacement / Total time
Initial position = 0 m
Final position = 70 m
Total displacement = 70 β 0
= 70 m
Total time = 4 + 2 + 3
= 9 s
Average velocity = 70 / 9
β 7.78 m/s
Therefore, the correct answer is 7.78 m/s.
Important concept:
The 2 s during which the particle remains at rest must still be included in the total time. Average velocity depends on total displacement and the complete elapsed time.
Initial position = 0 m
Final position = 70 m
Total displacement = 70 β 0
= 70 m
Total time = 4 + 2 + 3
= 9 s
Average velocity = 70 / 9
β 7.78 m/s
Therefore, the correct answer is 7.78 m/s.
Important concept:
The 2 s during which the particle remains at rest must still be included in the total time. Average velocity depends on total displacement and the complete elapsed time.
19
A particle moves along a straight line with position x = (2tΒ² + 3t + 5) m, where t is in seconds. What is the average velocity during the interval from t = 1 s to t = 4 s?
Practice
A
11 m/s
B
13 m/s
C
15 m/s
D
17 m/s
β Show Answer
β Correct Answer:
B
(13 m/s)
π‘ Explanation
Average velocity = [x(tβ) β x(tβ)] / (tβ β tβ)
Given:
x = 2tΒ² + 3t + 5
At t = 1 s:
xβ = 2(1)Β² + 3(1) + 5
= 2 + 3 + 5
= 10 m
At t = 4 s:
xβ = 2(4)Β² + 3(4) + 5
= 32 + 12 + 5
= 49 m
Displacement = 49 β 10
= 39 m
Time interval = 4 β 1
= 3 s
Average velocity = 39 / 3
= 13 m/s
Therefore, the correct answer is 13 m/s.
Given:
x = 2tΒ² + 3t + 5
At t = 1 s:
xβ = 2(1)Β² + 3(1) + 5
= 2 + 3 + 5
= 10 m
At t = 4 s:
xβ = 2(4)Β² + 3(4) + 5
= 32 + 12 + 5
= 49 m
Displacement = 49 β 10
= 39 m
Time interval = 4 β 1
= 3 s
Average velocity = 39 / 3
= 13 m/s
Therefore, the correct answer is 13 m/s.
20
A particle moves along a straight line. Its velocity is 10 m/s for the first 4 s and β6 m/s for the next 2 s. What is its average velocity during the entire 6 s?
Practice
A
4 m/s
B
4.67 m/s
C
6 m/s
D
8 m/s
β Show Answer
β Correct Answer:
B
(4.67 m/s)
π‘ Explanation
Average velocity = Total displacement / Total time
For the first 4 s:
Velocity = +10 m/s
Displacement = velocity Γ time
= 10 Γ 4
= 40 m
For the next 2 s:
Velocity = β6 m/s
Displacement = β6 Γ 2
= β12 m
Total displacement = 40 β 12
= 28 m
Total time = 4 + 2
= 6 s
Average velocity = 28 / 6
= 14/3
β 4.67 m/s
Therefore, the correct answer is 4.67 m/s.
Important concept:
Average velocity is NOT the ordinary average of the two velocities because the two velocities act for different time intervals. We must use total displacement divided by total time.
For the first 4 s:
Velocity = +10 m/s
Displacement = velocity Γ time
= 10 Γ 4
= 40 m
For the next 2 s:
Velocity = β6 m/s
Displacement = β6 Γ 2
= β12 m
Total displacement = 40 β 12
= 28 m
Total time = 4 + 2
= 6 s
Average velocity = 28 / 6
= 14/3
β 4.67 m/s
Therefore, the correct answer is 4.67 m/s.
Important concept:
Average velocity is NOT the ordinary average of the two velocities because the two velocities act for different time intervals. We must use total displacement divided by total time.
21
A particle starts from x = 0 m. During the first 5 s, its velocity increases uniformly from 2 m/s to 8 m/s. What is its average velocity during these 5 s?
Practice
A
3 m/s
B
5 m/s
C
6 m/s
D
8 m/s
β Show Answer
β Correct Answer:
B
(5 m/s)
π‘ Explanation
Since the velocity changes uniformly with time, the average velocity is the arithmetic mean of the initial and final velocities.
Initial velocity, u = 2 m/s
Final velocity, v = 8 m/s
Average velocity = (u + v) / 2
= (2 + 8) / 2
= 10 / 2
= 5 m/s
Therefore, the correct answer is 5 m/s.
Important concept:
For uniformly changing velocity,
Average velocity = (Initial velocity + Final velocity) / 2.
This shortcut is valid only when the velocity changes uniformly.
Initial velocity, u = 2 m/s
Final velocity, v = 8 m/s
Average velocity = (u + v) / 2
= (2 + 8) / 2
= 10 / 2
= 5 m/s
Therefore, the correct answer is 5 m/s.
Important concept:
For uniformly changing velocity,
Average velocity = (Initial velocity + Final velocity) / 2.
This shortcut is valid only when the velocity changes uniformly.
22
A particle moves along a straight line. It covers 20 m in the first 2 s and 50 m in the next 5 s, both in the same direction. What is its average velocity for the entire 7 s?
Practice
A
7 m/s
B
10 m/s
C
12 m/s
D
14 m/s
β Show Answer
β Correct Answer:
B
(10 m/s)
π‘ Explanation
Average velocity = Total displacement / Total time
Since both parts are in the same direction, displacement equals the distance travelled.
First displacement = 20 m
Second displacement = 50 m
Total displacement = 20 + 50
= 70 m
Total time = 2 + 5
= 7 s
Average velocity = 70 / 7
= 10 m/s
Therefore, the correct answer is 10 m/s.
Since both parts are in the same direction, displacement equals the distance travelled.
First displacement = 20 m
Second displacement = 50 m
Total displacement = 20 + 50
= 70 m
Total time = 2 + 5
= 7 s
Average velocity = 70 / 7
= 10 m/s
Therefore, the correct answer is 10 m/s.
23
A particle moves along the x-axis according to x = tΒ³ β 6tΒ² + 9t + 4, where x is in metres and t is in seconds. What is the average velocity between t = 1 s and t = 4 s?
Practice
A
β3 m/s
B
0 m/s
C
3 m/s
D
6 m/s
β Show Answer
β Correct Answer:
B
(0 m/s)
π‘ Explanation
Average velocity = [x(tβ) β x(tβ)] / (tβ β tβ)
Given:
x = tΒ³ β 6tΒ² + 9t + 4
At t = 1 s:
xβ = (1)Β³ β 6(1)Β² + 9(1) + 4
= 1 β 6 + 9 + 4
= 8 m
At t = 4 s:
xβ = (4)Β³ β 6(4)Β² + 9(4) + 4
= 64 β 96 + 36 + 4
= 8 m
Displacement = xβ β xβ
= 8 β 8
= 0 m
Time interval = 4 β 1
= 3 s
Average velocity = 0 / 3
= 0 m/s
Therefore, the correct answer is 0 m/s.
Important concept:
The particle may move significantly during the interval, but if its final position is the same as its initial position, its average velocity is zero. Average velocity depends only on net displacement and total time.
Given:
x = tΒ³ β 6tΒ² + 9t + 4
At t = 1 s:
xβ = (1)Β³ β 6(1)Β² + 9(1) + 4
= 1 β 6 + 9 + 4
= 8 m
At t = 4 s:
xβ = (4)Β³ β 6(4)Β² + 9(4) + 4
= 64 β 96 + 36 + 4
= 8 m
Displacement = xβ β xβ
= 8 β 8
= 0 m
Time interval = 4 β 1
= 3 s
Average velocity = 0 / 3
= 0 m/s
Therefore, the correct answer is 0 m/s.
Important concept:
The particle may move significantly during the interval, but if its final position is the same as its initial position, its average velocity is zero. Average velocity depends only on net displacement and total time.
24
The position of a particle moving along the x-axis is given by x = 12t β 2tΒ², where x is in metres and t is in seconds. At what time interval is the average velocity equal to zero?
Practice
A
From 0 s to 3 s
B
From 0 s to 6 s
C
From 2 s to 6 s
D
From 3 s to 6 s
β Show Answer
β Correct Answer:
B
(From 0 s to 6 s)
π‘ Explanation
Average velocity = [x(tβ) β x(tβ)] / (tβ β tβ)
For average velocity to be zero,
x(tβ) = x(tβ).
Given:
x = 12t β 2tΒ²
At t = 0 s:
x(0) = 0
At t = 6 s:
x(6) = 12(6) β 2(6)Β²
= 72 β 72
= 0 m
Thus,
x(6) = x(0)
Therefore,
Average velocity from 0 s to 6 s
= [0 β 0] / (6 β 0)
= 0 m/s
Hence, the correct answer is from 0 s to 6 s.
Important JEE concept:
Zero average velocity does NOT mean that the particle remained at rest. It means only that the net displacement during the chosen time interval is zero.
Here the particle moves away from x = 0, reaches a maximum position, and later returns to x = 0.
For average velocity to be zero,
x(tβ) = x(tβ).
Given:
x = 12t β 2tΒ²
At t = 0 s:
x(0) = 0
At t = 6 s:
x(6) = 12(6) β 2(6)Β²
= 72 β 72
= 0 m
Thus,
x(6) = x(0)
Therefore,
Average velocity from 0 s to 6 s
= [0 β 0] / (6 β 0)
= 0 m/s
Hence, the correct answer is from 0 s to 6 s.
Important JEE concept:
Zero average velocity does NOT mean that the particle remained at rest. It means only that the net displacement during the chosen time interval is zero.
Here the particle moves away from x = 0, reaches a maximum position, and later returns to x = 0.
25
The position of a particle moving along the x-axis is given by x = tΒ³ β 3tΒ² + 2t + 5, where x is in metres and t is in seconds. What is the average velocity of the particle during the interval in which it moves from x = 5 m at t = 0 s to the next instant when it again reaches x = 5 m?
Practice
A
0 m/s
B
1 m/s
C
2 m/s
D
3 m/s
β Show Answer
β Correct Answer:
A
(0 m/s)
π‘ Explanation
Average velocity = Total displacement / Total time
Given:
x = tΒ³ β 3tΒ² + 2t + 5
Initially, at t = 0:
x(0) = 5 m
We need the next time when x becomes 5 m.
Therefore,
tΒ³ β 3tΒ² + 2t + 5 = 5
tΒ³ β 3tΒ² + 2t = 0
t(tΒ² β 3t + 2) = 0
t(t β 1)(t β 2) = 0
The possible times are:
t = 0 s, 1 s, 2 s
Since t = 0 s is the initial instant, the next instant is:
t = 1 s
At both t = 0 s and t = 1 s, the position is 5 m.
Therefore,
Total displacement = 5 β 5
= 0 m
Total time = 1 β 0
= 1 s
Average velocity = 0 / 1
= 0 m/s
Therefore, the correct answer is 0 m/s.
Important JEE concept:
When a particle returns to its initial position, its average velocity over that interval is zero, regardless of the distance travelled in between.
Given:
x = tΒ³ β 3tΒ² + 2t + 5
Initially, at t = 0:
x(0) = 5 m
We need the next time when x becomes 5 m.
Therefore,
tΒ³ β 3tΒ² + 2t + 5 = 5
tΒ³ β 3tΒ² + 2t = 0
t(tΒ² β 3t + 2) = 0
t(t β 1)(t β 2) = 0
The possible times are:
t = 0 s, 1 s, 2 s
Since t = 0 s is the initial instant, the next instant is:
t = 1 s
At both t = 0 s and t = 1 s, the position is 5 m.
Therefore,
Total displacement = 5 β 5
= 0 m
Total time = 1 β 0
= 1 s
Average velocity = 0 / 1
= 0 m/s
Therefore, the correct answer is 0 m/s.
Important JEE concept:
When a particle returns to its initial position, its average velocity over that interval is zero, regardless of the distance travelled in between.
26
A particle moves along the x-axis with position x = tΒ³ β 6tΒ² + 12t, where x is in metres and t is in seconds. Find the average velocity of the particle from t = 1 s to t = 5 s.
Practice
A
4 m/s
B
6 m/s
C
7 m/s
D
10 m/s
β Show Answer
β Correct Answer:
C
(7 m/s)
π‘ Explanation
Average velocity = [x(tβ) β x(tβ)] / (tβ β tβ)
Given:
x = tΒ³ β 6tΒ² + 12t
At t = 1 s:
xβ = (1)Β³ β 6(1)Β² + 12(1)
= 1 β 6 + 12
= 7 m
At t = 5 s:
xβ = (5)Β³ β 6(5)Β² + 12(5)
= 125 β 150 + 60
= 35 m
Displacement = xβ β xβ
= 35 β 7
= 28 m
Time interval = 5 β 1
= 4 s
Average velocity = 28 / 4
= 7 m/s
Therefore, the correct answer is 7 m/s.
Given:
x = tΒ³ β 6tΒ² + 12t
At t = 1 s:
xβ = (1)Β³ β 6(1)Β² + 12(1)
= 1 β 6 + 12
= 7 m
At t = 5 s:
xβ = (5)Β³ β 6(5)Β² + 12(5)
= 125 β 150 + 60
= 35 m
Displacement = xβ β xβ
= 35 β 7
= 28 m
Time interval = 5 β 1
= 4 s
Average velocity = 28 / 4
= 7 m/s
Therefore, the correct answer is 7 m/s.
27
A particle moves along a straight line. It travels with velocities of 8 m/s, 14 m/s and 20 m/s during three successive time intervals of 4 s each. What is the average velocity for the entire motion?
Practice
A
12 m/s
B
14 m/s
C
16 m/s
D
18 m/s
β Show Answer
β Correct Answer:
B
(14 m/s)
π‘ Explanation
Since the time intervals are equal, the average velocity is the arithmetic mean of the individual velocities.
Average velocity = (vβ + vβ + vβ) / 3
Given:
vβ = 8 m/s
vβ = 14 m/s
vβ = 20 m/s
Therefore,
Average velocity = (8 + 14 + 20) / 3
= 42 / 3
= 14 m/s
Therefore, the correct answer is 14 m/s.
Important concept:
When equal time intervals are involved, average velocity can be found directly using the arithmetic mean of the velocities.
Average velocity = (vβ + vβ + vβ) / 3
Given:
vβ = 8 m/s
vβ = 14 m/s
vβ = 20 m/s
Therefore,
Average velocity = (8 + 14 + 20) / 3
= 42 / 3
= 14 m/s
Therefore, the correct answer is 14 m/s.
Important concept:
When equal time intervals are involved, average velocity can be found directly using the arithmetic mean of the velocities.
28
A particle moves along a straight line with velocities 10 m/s for 2 s, 20 m/s for 3 s, and 30 m/s for 5 s. What is its average velocity for the entire motion?
Practice
A
20 m/s
B
21 m/s
C
23 m/s
D
25 m/s
β Show Answer
β Correct Answer:
B
(21 m/s)
π‘ Explanation
Since the time intervals are unequal, we cannot use the simple arithmetic mean of the velocities.
Average velocity = Total displacement / Total time
Displacement in first interval:
sβ = vβtβ
= 10 Γ 2
= 20 m
Displacement in second interval:
sβ = vβtβ
= 20 Γ 3
= 60 m
Displacement in third interval:
sβ = vβtβ
= 30 Γ 5
= 150 m
Total displacement = 20 + 60 + 150
= 230 m
Total time = 2 + 3 + 5
= 10 s
Average velocity = 230 / 10
= 23 m/s
Therefore, the correct answer is 23 m/s.
Important concept:
For unequal time intervals, average velocity is the time-weighted average:
v_avg = (vβtβ + vβtβ + vβtβ) / (tβ + tβ + tβ).
Average velocity = Total displacement / Total time
Displacement in first interval:
sβ = vβtβ
= 10 Γ 2
= 20 m
Displacement in second interval:
sβ = vβtβ
= 20 Γ 3
= 60 m
Displacement in third interval:
sβ = vβtβ
= 30 Γ 5
= 150 m
Total displacement = 20 + 60 + 150
= 230 m
Total time = 2 + 3 + 5
= 10 s
Average velocity = 230 / 10
= 23 m/s
Therefore, the correct answer is 23 m/s.
Important concept:
For unequal time intervals, average velocity is the time-weighted average:
v_avg = (vβtβ + vβtβ + vβtβ) / (tβ + tβ + tβ).
29
A car travels the first half of a journey at 30 km/h and the second half at 50 km/h. What is its average velocity for the entire journey?
Practice
A
35 km/h
B
37.5 km/h
C
40 km/h
D
42.5 km/h
β Show Answer
β Correct Answer:
B
(37.5 km/h)
π‘ Explanation
Since the car covers equal distances at the two velocities, the average velocity is given by the harmonic mean.
Average velocity = (2vβvβ) / (vβ + vβ)
Given:
vβ = 30 km/h
vβ = 50 km/h
Therefore,
Average velocity = (2 Γ 30 Γ 50) / (30 + 50)
= 3000 / 80
= 37.5 km/h
Therefore, the correct answer is 37.5 km/h.
Important concept:
For equal distances, average velocity is NOT the arithmetic mean.
It is the harmonic mean:
v_avg = 2vβvβ / (vβ + vβ).
Average velocity = (2vβvβ) / (vβ + vβ)
Given:
vβ = 30 km/h
vβ = 50 km/h
Therefore,
Average velocity = (2 Γ 30 Γ 50) / (30 + 50)
= 3000 / 80
= 37.5 km/h
Therefore, the correct answer is 37.5 km/h.
Important concept:
For equal distances, average velocity is NOT the arithmetic mean.
It is the harmonic mean:
v_avg = 2vβvβ / (vβ + vβ).
30
The velocity of a particle increases uniformly from 0 m/s to 20 m/s in 5 s and then decreases uniformly to 10 m/s in the next 5 s. What is the average velocity of the particle during the entire 10 s?
Practice
A
10 m/s
B
12.5 m/s
C
15 m/s
D
20 m/s
β Show Answer
β Correct Answer:
B
(12.5 m/s)
π‘ Explanation
Average velocity = Total displacement / Total time
Since velocity changes uniformly in each interval, calculate the displacement using the area under the velocity-time graph.
First 5 s:
Initial velocity = 0 m/s
Final velocity = 20 m/s
Displacement in first 5 s:
sβ = [(u + v) / 2] Γ t
= [(0 + 20) / 2] Γ 5
= 10 Γ 5
= 50 m
Next 5 s:
Initial velocity = 20 m/s
Final velocity = 10 m/s
Displacement in next 5 s:
sβ = [(20 + 10) / 2] Γ 5
= 15 Γ 5
= 75 m
Total displacement = 50 + 75
= 125 m
Total time = 5 + 5
= 10 s
Average velocity = 125 / 10
= 12.5 m/s
Therefore, the correct answer is 12.5 m/s.
Important concept:
Average velocity can be found from a velocity-time graph using:
Average velocity = Area under v-t graph / Total time.
Since velocity changes uniformly in each interval, calculate the displacement using the area under the velocity-time graph.
First 5 s:
Initial velocity = 0 m/s
Final velocity = 20 m/s
Displacement in first 5 s:
sβ = [(u + v) / 2] Γ t
= [(0 + 20) / 2] Γ 5
= 10 Γ 5
= 50 m
Next 5 s:
Initial velocity = 20 m/s
Final velocity = 10 m/s
Displacement in next 5 s:
sβ = [(20 + 10) / 2] Γ 5
= 15 Γ 5
= 75 m
Total displacement = 50 + 75
= 125 m
Total time = 5 + 5
= 10 s
Average velocity = 125 / 10
= 12.5 m/s
Therefore, the correct answer is 12.5 m/s.
Important concept:
Average velocity can be found from a velocity-time graph using:
Average velocity = Area under v-t graph / Total time.
31
A particle moves along a straight line without changing direction. It covers 120 m at 12 m/s, 180 m at 18 m/s, and 300 m at 15 m/s. What is its average velocity for the entire journey?
Practice
A
14 m/s
B
15 m/s
C
16 m/s
D
18 m/s
β Show Answer
β Correct Answer:
B
(15 m/s)
π‘ Explanation
Since the distances and times are unequal, use the general definition:
Average velocity = Total displacement / Total time
Step 1: Calculate the time for each segment.
tβ = sβ / vβ
= 120 / 12
= 10 s
tβ = sβ / vβ
= 180 / 18
= 10 s
tβ = sβ / vβ
= 300 / 15
= 20 s
Step 2: Calculate total displacement.
Since the particle does not turn back,
Total displacement = 120 + 180 + 300
= 600 m
Step 3: Calculate total time.
Total time = 10 + 10 + 20
= 40 s
Step 4: Calculate average velocity.
Average velocity = 600 / 40
= 15 m/s
Therefore, the correct answer is 15 m/s.
Important concept:
For unequal distances and unequal time intervals, always use:
Average velocity = Total displacement / Total time.
Average velocity = Total displacement / Total time
Step 1: Calculate the time for each segment.
tβ = sβ / vβ
= 120 / 12
= 10 s
tβ = sβ / vβ
= 180 / 18
= 10 s
tβ = sβ / vβ
= 300 / 15
= 20 s
Step 2: Calculate total displacement.
Since the particle does not turn back,
Total displacement = 120 + 180 + 300
= 600 m
Step 3: Calculate total time.
Total time = 10 + 10 + 20
= 40 s
Step 4: Calculate average velocity.
Average velocity = 600 / 40
= 15 m/s
Therefore, the correct answer is 15 m/s.
Important concept:
For unequal distances and unequal time intervals, always use:
Average velocity = Total displacement / Total time.
32
A particle covers the first half of its total journey distance at 15 m/s. The remaining half is covered in two equal time intervals with speeds 6 m/s and 10 m/s. What is the average velocity for the entire journey?
Practice
A
9 m/s
B
10.43 m/s
C
11 m/s
D
12 m/s
β Show Answer
β Correct Answer:
B
(10.43 m/s)
π‘ Explanation
Step 1: Find the effective velocity for the second half.
The second half is covered in equal time intervals.
Therefore, its average velocity is the arithmetic mean:
v' = (vβ + vβ) / 2
= (6 + 10) / 2
= 16 / 2
= 8 m/s
Step 2: Now the journey consists of two equal-distance halves.
First half velocity = 15 m/s
Second half effective velocity = 8 m/s
For equal distances, use the harmonic mean:
v_avg = 2vβv' / (vβ + v')
= (2 Γ 15 Γ 8) / (15 + 8)
= 240 / 23
β 10.43 m/s
Therefore, the average velocity is approximately 10.43 m/s.
Important concept:
This is a hybrid model. First reduce the equal-time portion using the arithmetic mean, then combine the resulting effective velocity with the equal-distance portion using the harmonic-mean formula.
The second half is covered in equal time intervals.
Therefore, its average velocity is the arithmetic mean:
v' = (vβ + vβ) / 2
= (6 + 10) / 2
= 16 / 2
= 8 m/s
Step 2: Now the journey consists of two equal-distance halves.
First half velocity = 15 m/s
Second half effective velocity = 8 m/s
For equal distances, use the harmonic mean:
v_avg = 2vβv' / (vβ + v')
= (2 Γ 15 Γ 8) / (15 + 8)
= 240 / 23
β 10.43 m/s
Therefore, the average velocity is approximately 10.43 m/s.
Important concept:
This is a hybrid model. First reduce the equal-time portion using the arithmetic mean, then combine the resulting effective velocity with the equal-distance portion using the harmonic-mean formula.
33
A particle moves in a straight line with uniform acceleration. Its velocity increases from 8 m/s to 32 m/s in 6 s. What is its average velocity during this 6-second interval?
Practice
A
16 m/s
B
18 m/s
C
20 m/s
D
24 m/s
β Show Answer
β Correct Answer:
C
(20 m/s)
π‘ Explanation
For uniformly accelerated motion, the average velocity is the arithmetic mean of the initial and final velocities.
v_avg = (u + v) / 2
Given:
u = 8 m/s
v = 32 m/s
Therefore,
v_avg = (8 + 32) / 2
= 40 / 2
= 20 m/s
Therefore, the correct answer is 20 m/s.
Important concept:
For uniformly accelerated motion, average velocity depends only on the initial and final velocities:
v_avg = (u + v) / 2
The duration of the interval is not needed once u and v are known.
v_avg = (u + v) / 2
Given:
u = 8 m/s
v = 32 m/s
Therefore,
v_avg = (8 + 32) / 2
= 40 / 2
= 20 m/s
Therefore, the correct answer is 20 m/s.
Important concept:
For uniformly accelerated motion, average velocity depends only on the initial and final velocities:
v_avg = (u + v) / 2
The duration of the interval is not needed once u and v are known.
34
The position of a particle moving along the x-axis is given by x(t) = 2tΒ² β 8t + 3, where x is in metres and t is in seconds. Find the magnitude of its average velocity between t = 1 s and t = 4 s.
Practice
A
2 m/s
B
4 m/s
C
6 m/s
D
8 m/s
β Show Answer
β Correct Answer:
B
(4 m/s)
π‘ Explanation
Average velocity = Displacement / Time interval
Step 1: Find the position at tβ = 1 s.
x(1) = 2(1)Β² β 8(1) + 3
= 2 β 8 + 3
= β3 m
Step 2: Find the position at tβ = 4 s.
x(4) = 2(4)Β² β 8(4) + 3
= 32 β 32 + 3
= 3 m
Step 3: Calculate displacement.
Ξx = x(4) β x(1)
= 3 β (β3)
= 6 m
Step 4: Calculate the time interval.
Ξt = 4 β 1
= 3 s
Step 5: Calculate average velocity.
v_avg = Ξx / Ξt
= 6 / 3
= 2 m/s
Therefore, the magnitude of average velocity is 2 m/s.
Important concept:
For a position function x(t), always calculate the positions at the two specified times first. Then use:
v_avg = [x(tβ) β x(tβ)] / (tβ β tβ).
Step 1: Find the position at tβ = 1 s.
x(1) = 2(1)Β² β 8(1) + 3
= 2 β 8 + 3
= β3 m
Step 2: Find the position at tβ = 4 s.
x(4) = 2(4)Β² β 8(4) + 3
= 32 β 32 + 3
= 3 m
Step 3: Calculate displacement.
Ξx = x(4) β x(1)
= 3 β (β3)
= 6 m
Step 4: Calculate the time interval.
Ξt = 4 β 1
= 3 s
Step 5: Calculate average velocity.
v_avg = Ξx / Ξt
= 6 / 3
= 2 m/s
Therefore, the magnitude of average velocity is 2 m/s.
Important concept:
For a position function x(t), always calculate the positions at the two specified times first. Then use:
v_avg = [x(tβ) β x(tβ)] / (tβ β tβ).
35
A particle moves along a circular path of radius 7 m with constant speed 10 m/s. It turns through an angle of 60Β°. What is the magnitude of its average velocity?
Practice
A
5.00 m/s
B
9.55 m/s
C
8.66 m/s
D
10.00 m/s
β Show Answer
β Correct Answer:
B
(9.55 m/s)
π‘ Explanation
Average velocity = Magnitude of displacement / Total time
For circular motion, the displacement is the chord joining the initial and final positions.
Step 1: Find the displacement.
For a circular path,
Displacement = 2R sin(ΞΈ/2)
Given:
R = 7 m
ΞΈ = 60Β°
Displacement = 2 Γ 7 Γ sin(30Β°)
= 14 Γ 1/2
= 7 m
Step 2: Find the time taken.
Distance travelled along the arc = RΞΈ
Convert ΞΈ into radians:
ΞΈ = 60Β° = Ο/3 rad
Arc length = 7 Γ Ο/3
= 7Ο/3 m
Speed = Distance / Time
Therefore,
Time = Distance / Speed
= (7Ο/3) / 10
= 7Ο/30 s
Step 3: Calculate average velocity.
Average velocity = Displacement / Time
= 7 / (7Ο/30)
= 30/Ο
β 9.55 m/s
Therefore, the magnitude of average velocity is approximately 9.55 m/s.
For circular motion, the displacement is the chord joining the initial and final positions.
Step 1: Find the displacement.
For a circular path,
Displacement = 2R sin(ΞΈ/2)
Given:
R = 7 m
ΞΈ = 60Β°
Displacement = 2 Γ 7 Γ sin(30Β°)
= 14 Γ 1/2
= 7 m
Step 2: Find the time taken.
Distance travelled along the arc = RΞΈ
Convert ΞΈ into radians:
ΞΈ = 60Β° = Ο/3 rad
Arc length = 7 Γ Ο/3
= 7Ο/3 m
Speed = Distance / Time
Therefore,
Time = Distance / Speed
= (7Ο/3) / 10
= 7Ο/30 s
Step 3: Calculate average velocity.
Average velocity = Displacement / Time
= 7 / (7Ο/30)
= 30/Ο
β 9.55 m/s
Therefore, the magnitude of average velocity is approximately 9.55 m/s.
36
A person takes 40 s to walk up a stationary escalator. When the person stands still, the moving escalator carries them to the top in 60 s. How long will it take the person to reach the top if they walk up the moving escalator at the same walking speed?
Practice
A
20 s
B
24 s
C
30 s
D
36 s
β Show Answer
β Correct Answer:
B
(24 s)
π‘ Explanation
Let the length of the escalator be L.
Step 1: Find the person's walking speed relative to the escalator.
When the escalator is stationary, the person covers the distance L in 40 s.
Walking speed = L / 40
Step 2: Find the escalator's speed.
When the person stands still, the escalator carries them through distance L in 60 s.
Escalator speed = L / 60
Step 3: When the person walks on the moving escalator, the speeds add.
Ground speed = Walking speed + Escalator speed
= L/40 + L/60
Taking the LCM of 40 and 60:
= 3L/120 + 2L/120
= 5L/120
= L/24
Step 4: Find the time taken.
Time = Distance / Speed
= L / (L/24)
= 24 s
Therefore, the correct answer is 24 s.
Important concept:
When two motions occur in the same direction, their velocities relative to the ground add. Here, the person's walking velocity and the escalator's velocity both act upward.
Step 1: Find the person's walking speed relative to the escalator.
When the escalator is stationary, the person covers the distance L in 40 s.
Walking speed = L / 40
Step 2: Find the escalator's speed.
When the person stands still, the escalator carries them through distance L in 60 s.
Escalator speed = L / 60
Step 3: When the person walks on the moving escalator, the speeds add.
Ground speed = Walking speed + Escalator speed
= L/40 + L/60
Taking the LCM of 40 and 60:
= 3L/120 + 2L/120
= 5L/120
= L/24
Step 4: Find the time taken.
Time = Distance / Speed
= L / (L/24)
= 24 s
Therefore, the correct answer is 24 s.
Important concept:
When two motions occur in the same direction, their velocities relative to the ground add. Here, the person's walking velocity and the escalator's velocity both act upward.
37
The position-time graph of a particle consists of three straight-line segments. From t = 0 s to t = 3 s, the particle moves from x = 2 m to x = 14 m. From t = 3 s to t = 5 s, it remains at x = 14 m. From t = 5 s to t = 9 s, it moves from x = 14 m to x = 6 m. Find the average velocity and average speed from t = 0 s to t = 9 s.
Practice
A
Average velocity = 4/9 m/s, Average speed = 20/9 m/s
B
Average velocity = 4/9 m/s, Average speed = 8/3 m/s
C
Average velocity = 8/9 m/s, Average speed = 20/9 m/s
D
Average velocity = 2/3 m/s, Average speed = 8/3 m/s
β Show Answer
β Correct Answer:
A
(Average velocity = 4/9 m/s, Average speed = 20/9 m/s)
π‘ Explanation
Step 1: Find the initial and final positions.
Initial position:
xα΅’ = 2 m
Final position:
x_f = 6 m
Total time:
Ξt = 9 β 0 = 9 s
Step 2: Calculate average velocity.
Average velocity = (x_f β xα΅’) / (t_f β tα΅’)
= (6 β 2) / 9
= 4/9 m/s
Step 3: Calculate total distance travelled.
First segment:
Distance = |14 β 2|
= 12 m
Second segment:
The particle remains at x = 14 m.
Distance = 0 m
Third segment:
Distance = |6 β 14|
= 8 m
Total distance = 12 + 0 + 8
= 20 m
Step 4: Calculate average speed.
Average speed = Total distance / Total time
= 20 / 9
β 2.22 m/s
Therefore:
Average velocity = 4/9 m/s
Average speed = 20/9 m/s
Important concept:
On an x-t graph, average velocity depends only on the initial and final positions, whereas average speed depends on the complete path travelled.
Initial position:
xα΅’ = 2 m
Final position:
x_f = 6 m
Total time:
Ξt = 9 β 0 = 9 s
Step 2: Calculate average velocity.
Average velocity = (x_f β xα΅’) / (t_f β tα΅’)
= (6 β 2) / 9
= 4/9 m/s
Step 3: Calculate total distance travelled.
First segment:
Distance = |14 β 2|
= 12 m
Second segment:
The particle remains at x = 14 m.
Distance = 0 m
Third segment:
Distance = |6 β 14|
= 8 m
Total distance = 12 + 0 + 8
= 20 m
Step 4: Calculate average speed.
Average speed = Total distance / Total time
= 20 / 9
β 2.22 m/s
Therefore:
Average velocity = 4/9 m/s
Average speed = 20/9 m/s
Important concept:
On an x-t graph, average velocity depends only on the initial and final positions, whereas average speed depends on the complete path travelled.
38
A particle has the following velocity-time motion: from t = 0 s to t = 3 s, its velocity increases uniformly from 0 to 12 m/s. From t = 3 s to t = 5 s, it moves with constant velocity 12 m/s. From t = 5 s to t = 9 s, its velocity decreases uniformly from 12 m/s to β4 m/s, crossing the time axis at t = 8 s. Calculate the average velocity and average speed during the 9-second interval.
Practice
A
Average velocity = 6.44 m/s, Average speed = 6.89 m/s
B
Average velocity = 6.89 m/s, Average speed = 6.44 m/s
C
Average velocity = 8 m/s, Average speed = 10 m/s
D
Average velocity = 10 m/s, Average speed = 8 m/s
β Show Answer
β Correct Answer:
A
(Average velocity = 6.44 m/s, Average speed = 6.89 m/s)
π‘ Explanation
The area under a velocity-time graph represents displacement. For average velocity, use the signed areas. For average speed, use the absolute areas.
Step 1: Calculate the areas of the different regions.
Area 1: Triangle from t = 0 s to t = 3 s
Base = 3 s
Height = 12 m/s
Aβ = (1/2) Γ 3 Γ 12
= 18 m
Area 2: Rectangle from t = 3 s to t = 5 s
Base = 2 s
Height = 12 m/s
Aβ = 2 Γ 12
= 24 m
Area 3: From t = 5 s to t = 8 s, velocity is positive.
This is a trapezium with velocities 12 m/s and 0 m/s.
Aβ = (1/2)(12 + 0) Γ 3
= 18 m
Area 4: From t = 8 s to t = 9 s, velocity is negative.
This is a triangle with base 1 s and height 4 m/s.
Aβ = β(1/2) Γ 1 Γ 4
= β2 m
Step 2: Calculate net displacement.
Net displacement = Aβ + Aβ + Aβ + Aβ
= 18 + 24 + 18 β 2
= 58 m
Step 3: Calculate average velocity.
Average velocity = Net displacement / Total time
= 58 / 9
β 6.44 m/s
Step 4: Calculate total distance.
Total distance = |Aβ| + |Aβ| + |Aβ| + |Aβ|
= 18 + 24 + 18 + 2
= 62 m
Average speed = Total distance / Total time
= 62 / 9
β 6.89 m/s
Therefore:
Average velocity β 6.44 m/s
Average speed β 6.89 m/s.
Important concept:
For a v-t graph, signed area gives displacement, while the sum of absolute areas gives total distance. Therefore, average velocity and average speed are generally different.
Step 1: Calculate the areas of the different regions.
Area 1: Triangle from t = 0 s to t = 3 s
Base = 3 s
Height = 12 m/s
Aβ = (1/2) Γ 3 Γ 12
= 18 m
Area 2: Rectangle from t = 3 s to t = 5 s
Base = 2 s
Height = 12 m/s
Aβ = 2 Γ 12
= 24 m
Area 3: From t = 5 s to t = 8 s, velocity is positive.
This is a trapezium with velocities 12 m/s and 0 m/s.
Aβ = (1/2)(12 + 0) Γ 3
= 18 m
Area 4: From t = 8 s to t = 9 s, velocity is negative.
This is a triangle with base 1 s and height 4 m/s.
Aβ = β(1/2) Γ 1 Γ 4
= β2 m
Step 2: Calculate net displacement.
Net displacement = Aβ + Aβ + Aβ + Aβ
= 18 + 24 + 18 β 2
= 58 m
Step 3: Calculate average velocity.
Average velocity = Net displacement / Total time
= 58 / 9
β 6.44 m/s
Step 4: Calculate total distance.
Total distance = |Aβ| + |Aβ| + |Aβ| + |Aβ|
= 18 + 24 + 18 + 2
= 62 m
Average speed = Total distance / Total time
= 62 / 9
β 6.89 m/s
Therefore:
Average velocity β 6.44 m/s
Average speed β 6.89 m/s.
Important concept:
For a v-t graph, signed area gives displacement, while the sum of absolute areas gives total distance. Therefore, average velocity and average speed are generally different.