Distance, Displacement, Average Speed, Average Velocity, Applications of Equations of Motion, Graphical Understanding

 

Physics Worksheet: Motion in One Dimension

Topics: Distance, Displacement, Average Speed, Average Velocity, Applications of Equations of Motion, Graphical Understanding

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Multiple Choice Questions

Q1.

A person walks 400 m east and then 300 m west. The total distance travelled is

A. 100 m

B. 300 m

C. 700 m

D. 400 m

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Q2.

For the motion in Question 1, the displacement is

A. 700 m east

B. 100 m east

C. 100 m west

D. 300 m west

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Q3.

An athlete runs 600 m in 120 s. The average speed is

A. 2 m/s

B. 4 m/s

C. 5 m/s

D. 6 m/s

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Q4.

A car travels 40 km north and then 30 km south in 2 hours. Its average velocity is

A. 35 km/h north

B. 5 km/h north

C. 20 km/h north

D. 15 km/h north

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Q5.

A body starts from rest and accelerates uniformly at 2 m/s² for 5 s. Its final velocity is

A. 5 m/s

B. 8 m/s

C. 10 m/s

D. 15 m/s

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Q6.

A body moves with initial velocity 5 m/s and acceleration 3 m/s² for 4 s. The final velocity is

A. 12 m/s

B. 15 m/s

C. 17 m/s

D. 20 m/s

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Q7.

A train accelerates uniformly from rest at 1.5 m/s² for 20 s. The distance travelled is

A. 150 m

B. 250 m

C. 300 m

D. 450 m

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Q8.

A car moving at 20 m/s comes to rest in 5 s. The acceleration is

A. −2 m/s²

B. −3 m/s²

C. −4 m/s²

D. −5 m/s²

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Q9.

A particle has

Initial velocity = 10 m/s

Acceleration = 2 m/s²

Time = 6 s

Its displacement is

A. 36 m

B. 72 m

C. 96 m

D. 120 m

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Q10.

Which quantity can never be negative?

A. Velocity

B. Acceleration

C. Speed

D. Displacement

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Q11.

The slope of a displacement-time graph represents

A. Acceleration

B. Distance

C. Velocity

D. Force

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Q12.

The area under a velocity-time graph represents

A. Speed

B. Displacement

C. Acceleration

D. Force

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Q13.

The slope of a velocity-time graph gives

A. Speed

B. Distance

C. Acceleration

D. Momentum

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Q14.

A horizontal line on a velocity-time graph indicates

A. Uniform acceleration

B. Zero velocity

C. Constant velocity

D. Increasing speed

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Q15.

A body travels equal distances with speeds 20 km/h and 30 km/h. Its average speed is

A. 24 km/h

B. 25 km/h

C. 26 km/h

D. 28 km/h

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Q16.

A ball is thrown vertically upward with velocity 20 m/s.

Its velocity at the highest point is

A. 20 m/s

B. 10 m/s

C. 5 m/s

D. 0 m/s

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Q17.

A cyclist rides around a circular track and returns to the starting point.

The displacement is

A. Equal to circumference

B. Radius

C. Diameter

D. Zero

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Q18.

A particle moves with constant acceleration.

Which graph is a straight line?

A. Position-Time

B. Velocity-Time

C. Distance-Time

D. Speed-Time (always)

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Q19.

A car increases its speed from 10 m/s to 30 m/s over a distance of 200 m.

The acceleration is

A. 1 m/s²

B. 2 m/s²

C. 3 m/s²

D. 4 m/s²

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Q20.

A train moving at 72 km/h has a speed in SI units equal to

A. 18 m/s

B. 20 m/s

C. 22 m/s

D. 25 m/s

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Answer Key

Q	Ans
1	C
2	B
3	C
4	B
5	C
6	C
7	C
8	C
9	C
10	C
11	C
12	B
13	C
14	C
15	A
16	D
17	D
18	B
19	B
20	B

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Detailed Step-by-Step Solutions

Q1

Distance = Total path travelled

= 400 + 300

= 700 m

Answer: C

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Q2

Displacement = 400 − 300

= 100 m east

Answer: B

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Q3

Average speed

$$=\frac{\text{Distance}}{\text{Time}} =\frac{600}{120}=5\text{ m/s}$$

Answer: C

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Q4

Displacement = 40 − 30 = 10 km north

Average velocity

$$=\frac{10}{2}=5\text{ km/h north}$$

Answer: B

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Q5

Using

$$v=u+at$$
$$=0+2(5)=10\text{ m/s}$$

Answer: C

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Q6

$$v=u+at$$
$$=5+3(4)=17\text{ m/s}$$

Answer: C

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Q7

Using

$$s=ut+\frac12 at^2$$

Since $u=0$,

$$=\frac12(1.5)(20)^2$$
$$=0.75\times400=300\text{ m}$$

Answer: C

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Q8

$$a=\frac{v-u}{t}$$ $$=\frac{0-20}{5} =-4\text{ m/s}^2$$

Answer: C

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Q9

$$s=ut+\frac12 at^2$$
$$=10(6)+\frac12(2)(36)$$
$$=60+36=96\text{ m}$$

Answer: C

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Q10

Speed is a scalar quantity and cannot be negative.

Answer: C

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Q11

Slope of displacement-time graph

$$=\frac{\Delta s}{\Delta t}$$

which is velocity.

Answer: C

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Q12

Area under velocity-time graph

$$=\text{Displacement}$$

Answer: B

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Q13

Slope of velocity-time graph

$$=\frac{\Delta v}{\Delta t}$$

which is acceleration.

Answer: C

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Q14

Horizontal line means velocity remains constant.

Answer: C

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Q15

For equal distances,

$$v_{\text{avg}} =\frac{2v_1v_2}{v_1+v_2}$$ $$=\frac{2(20)(30)}{20+30} =\frac{1200}{50} =24\text{ km/h}$$

We can also simply use direct formula, 

Average Speed = (d+d) / (t1 + t2) = 2d / (d/20 + d/30) = 2 x 20 x 30 / (20 + 30) = 24

Answer: A

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Q16

At the highest point,

Velocity = 0 m/s

Answer: D

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Q17

Initial position = Final position

Displacement = 0

Answer: D

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Q18

For constant acceleration,

Velocity changes uniformly.

Therefore velocity-time graph is a straight line.

Answer: B

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Q19

Using

$$v^2=u^2+2as$$
$$30^2=10^2+2(a)(200)$$
$$900=100+400a$$
$$800=400a$$
$$a=2\text{ m/s}^2$$

Answer: B

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Q20

Convert km/h to m/s

$$72\times\frac{1000}{3600} =20\text{ m/s}$$

Answer: B