MCQ Questions for Class 9 Physics Motion Quiz 6

If you were to throw a ball vertically upward with an initial velocity of

$50 \; ms^{-1}$ , approximately how long would it take for the ball to return to your hand? Assume air resistance is negligible. (Given

$g=10\; ms^{-2}$ )

0%
$2.5 \;s$
0%
$5.0 \;s$
0%
$7.5 \;s$
0%
$10 \;s$

Explanation

The only force acting on the ball is the force of gravity. The ball will ascend until gravity reduces its velocity to zero and then it will descend. Find the time it takes for the ball to reach its maximum height and then double the time to cover the round trip.
Using

$\displaystyle{ v }=u+at=u-gt$ , we get:

$0 { m }/{ s }=50 { m }/{ s }-\left( 10 { m }/{ { s }^{ 2 } } \right) t$
Therefore,

$t={ \left( 50 { m }/{ s } \right) }/{ \left( 10 { m }/{ { s }^{ 2 } } \right) }=5 s$
This is the time it takes the ball to reach its maximum height. The total round trip time is

$2t=10s$ .

A stone thrown upward with a speed

$u$ from the top of the tower reaches the ground with a velocity

$3u$ . The height of the tower is

0%
$3 { { u }^{ 2 } }/{ g }$
0%
$4 { { u }^{ 2 } }/{ g }$
0%
$6 { { u }^{ 2 } }/{ g }$
0%
$9 { { u }^{ 2 } }/{ g }$

Explanation

The stone rises up till its vertical velocity is zero and again reached the top of the tower with a speed

$u$ (downward). The speed of the stone at the base is

$3u$
WKT,

$v^{2} = u^{2} + 2gh$
where,

$v =$ final velocity

$u = -u$ initial velocity, -ve because it is thrown upward against gravity

$a =$ acceleration due to gravity

$h =$ height or distance

$\Rightarrow \left ( 3u \right )^{2} = \left ( -u \right )^{2} + 2gh$

$\Rightarrow 9u^2 - u^2 = 2gh$

$\Rightarrow h = \dfrac{4u^{2}}{g}$

If a particle moves in a circle with constant speed, its velocity :

0%
remains constant
0%
changes in magnitude
0%
changes direction
0%
changes both in magnitude and directions

A rubber ball is dropped from a height of

$5 m$ on a planet where the acceleration due to gravity is same as that on the surface of the earth. On bouncing, it rises to a height of

$1.8 m$ . the ball loses its velocity by a factor of

0%
$\displaystyle\frac { 3 }{ 5 }$
0%
$\displaystyle\frac { 9 }{ 25 }$
0%
$\displaystyle\frac { 2 }{ 5 }$
0%
$\displaystyle\frac { 16 }{ 25 }$

Explanation

Given,

Height from which ball is dropped

$s=5\ m$

Height attend by ball after bouncing

$s'=1.8\ m$

Gravitational acceleration

$g=9.8\ m/s^2$

We know by third equation of kinetic motion,

$v^{2} - u^{2} = 2gs$
where

$v =$ final velocity ,

$u =$ initial velocity ,

$S =$ distance
Now, for downward motion,

$u = 0$

$v^{2} - 0 = 2 \times 9.8 \times 5$

$\Rightarrow v = \sqrt{98} = 9.9$

Also for upward motion,

$v=0$

$0^{2} - u^{2} = 2 \times (-9.8) \times 1.8$

$\Rightarrow u = \sqrt{3528} = 5.94$

Fractional loss

$=\dfrac{9.9-5.94}{9.9} = 0.4$ or

$\dfrac25$

Option C

A steel ball is bouncing up and down on a steel plate with a period of oscillation 1 second. If

$g=10 { m }{ { s }^{-2 } }$ , then it bounces up to a height of

0%
$5$ $m$
0%
$10$ $m$
0%
$2.5$ $m$
0%
$1.25$ $m$

Explanation

The ball takes 1 second to bounce up and down through the same height; so, the time of fall is equal to time of rise i.e.

$\displaystyle\frac { 1 }{ 2 }$ second.

Using second law of motion,

$s= ut + \cfrac{1}{2}at^2$

Height,

$s=h,$ initial velocity,

$u=0$

$h=0 + \displaystyle\frac { 1 }{ 2 } g{ \left( \displaystyle\frac { 1 }{ 2 } \right) }^{ 2 }$

$h=\displaystyle\frac { 10 }{ 8 }$

$h=1.25 m$

So, the ball bounces up to a height of

$1.25m$ .

A body is projected vertically upward with an initial velocity

$u$ . If acceleration due to gravity is

$g$ , the time for which it remains in air, is

0%
$\displaystyle\frac{u}{g}$
0%
$ug$
0%
$\displaystyle\frac{2u}{g}$
0%
$\displaystyle\frac{u}{2g}$

Explanation

$v=u+gt$
at highest distance,v

$=$ 0,and g acts in opposite direction.
So,

$0=u-gt$

$t= \dfrac{u}{g}$ since time taken to ascent and descent is same.
So, the total time(t) is

$\dfrac{2u}{g}$ for which it remain in air.

A pebble is thrown vertically upwards with a speed of

$20\ ms^{-1}$ . How high will it be after

$2\ s$ ? (Take

$g=10\ ms^{-2}$ )

0%
40 m
0%
10 m
0%
20 m
0%
25 m

Explanation

Initial velocity ,

$u= 20\ m/s$
Since, the ball is thrown upward, so 2nd equation of motion will be given by,

$S = ut - \dfrac{1}{2}gt^{2}$

$\Rightarrow S = 20\times 2 - \dfrac{1}{2}\times 10\times 2^{2} = 40-20 =20\ m$

A body falls from the top of a building and reaches the ground

$2.5 s$ later. How high is the building? (Take

$g=10 m {s}^{-2}$ )

0%
30.6 m
0%
31.25 m
0%
30 m
0%
25 m

Explanation

Height of the building is given by 2nd equation of motion,

$S = ut + \dfrac{1}{2}gt^{2}$
now since initial velocity is zero so 2nd equation of motion, become,

$S = \dfrac{1}{2}gt^{2} = \dfrac{1}{2} \times 10 \times 2.5^{2} = 31.25 m$

Statement 1:Two balls of different masses are thrown vertically upward with same speed.They will pass through their point of projection in downward direction with the same speed.
Statement 2:The maximum height and downward velocity attained at the point of projection are independent of the mass of the ball

0%
Statement -1 is false , Statement- 2 is true.
0%
Statement-1 is true, Statement-2 is true , Statement-2 is a correct explanation for statement-1
0%
Statement-1 is true , Statement-2 is true; Statement-2 is not a correct explanation for statement-1
0%
Statement -1 is true, Statement-2 is false.

Explanation

Take upward direction as positive

Height is given by

$h=ut-\frac{1}{2}gt^2$

Velocity is given by

$v^2=u^2-2gh$
These equation are independent of mass.

A pebble thrown vertically upwards with an initial velocity

$50 \ m s^{-1}$ comes to a stop in 5 s. Find the retardation assuming displacement in upwards direction to be positive.

0%
$-10 \ m s^{-2}$
0%
$10 \ m s^{-2}$
0%
$250 \ m s^{-2}$
0%
$-250 \ m s^{-2}$

Explanation

Acceleration

$a=\dfrac{v-u}{t}= \dfrac{0-50 m s^{-1}}{5 s}$

$= -10 m s ^{-2}$

$\text{Retardation}=-\text{Acceleration}$

$=10\ ms^{-2}$

The velocity of an object increases at a constant rate from

$20 m s^{-1}$ to

$50 m s^{-1}$ in 10 s. Find the acceleration.

0%
$-3 \,m \,s^{-2}$
0%
$-30 \,m \,s^{-2}$
0%
$30 \,m \,s^{-2}$
0%
$3 \,m \,s^{-2}$

Explanation

Acceleration

$a=\dfrac {v-u}{t}=\dfrac{50 \ m s^{-1}-20 \ m s^{-1}}{10 s}$
or

$a=\dfrac{30 \ m s^{-1}}{10 s}$

$= 3 \ m s^{-2}$

A stone is dropped freely from the top of a tower and it reaches the ground in

$4\ s$ . Taking

$g=10\ ms^{-2}$ , calculate the height of the tower.

0%
80 m
0%
40 m
0%
4 m
0%
20 m

Explanation

$S = ut + \dfrac{1}{2}gt^{2}$
as

$u = 0$
So,

$S = \dfrac{1}{2}gt^{2} = \dfrac{1}{2} \times 10 \times 4^{2} = 80 m$

A truck of mass

$5\times 10^3\ kg$ starting from rest travels a distance of

$0.5\ km$ in

$10\ s$ when a force is applied on it. Calculate the acceleration acquired by the truck

0%
$25ms^{-2}$
0%
$1ms^{-2}$
0%
$10cms^{-2}$
0%
$10ms^{-2}$

Explanation

Here truck has initial velocity of 0 m/sec and travelled a distance of 0.5 Km which is also equal to 500 m in 10 seconds.

$s = ut+0.5at^{2}$

$s =0\times 10 + 0.5\times a\times 10^{2}$

$500=0.5\times a\times 100$

$a=10m/s^2$
Hence acceleration is

$10m/s^2$ .

An object falling freely from rest reaches ground in

$2 s$ . If acceleration due to gravity is

$9.8 ms^{-2}$ , the velocity of object on reaching the ground will be

0%
$9.8$ $m$ ${s}^{-1}$
0%
$4.9$ $m$ ${s}^{-1}$
0%
$19.6$ $m$ ${s}^{-1}$
0%
zero

Explanation

$v=u+gt$ , initially u

$=$ 0 and g acting in the direction of movement of body.

$v=0+9.8 \times 2$

$v=19.6 m/s$

A body initially at rest travels a distance 100 m in 5 s with a constant acceleration. Find the acceleration.

0%
$20 m s^{-2}$
0%
$4 m s^{-2}$
0%
$8 m s^{-2}$
0%
$10 m s^{-2}$

Explanation

From equation of motion

$S = ut + \dfrac{1}{2}at^2$

$100=0 \times 5 + \dfrac{1}{2} \times a \times (5)^2$
or

$100=\dfrac{1}{2} \times 25 a$
or
Acceleration

$a=\dfrac{100 \times 2}{25}= 8 m s^{-2}$

A body starts from rest with a uniform acceleration

$2 m s^{-2}$ . Find the distance covered by the body in

$2\ s$ .

0%
$4\ m$
0%
$2 \ m$
0%
$1 \ m$
0%
$0 \ m$

Explanation

$u = 0, a = 2 m/s^2, t = 2 s$
Putting the values in

$S = ut+\dfrac{1}{2}at^2$

$S = 0\times 2+\dfrac{1}{2}\times 2\times 2^2$

$S = 4 m$

Figure shows displacement-time graph of two vehicles A and B moving along a straight road. Which vehicle is moving faster?

0%
Vehicle A
0%
Vehicle B
0%
Vehicle A = Vehicle B
0%
None of the above

Give an example of motion in which average speed is not zero, but the average velocity is zero.

0%
A car moving with an average speed returns to the initial position
0%
A car at rest
0%
A car with non zero displacement.
0%
All of the above.

A body is moving vertically upwards. Its velocity changes at a constant rate from

$50 m s^{-1}$ to

$20 m s^{-1}$ in 3 s. What is its acceleration ?

0%
$-30 m s^{-2}$
0%
$-10 m s^{-2}$
0%
$10 m s^{-2}$
0%
$30 m s^{-2}$

Explanation

Acceleration,

$a = \dfrac{Final Velocity-Initial Velocity}{Time} = \dfrac{20 m/s-50 m/s}{3s} = -10 m/s^2$

A body starts with an initial velocity of

$10 m s^{-1}$ and acceleration

$5 m s^{-2}$ . Find the distance covered by it in

$5\ s$ .

0%
$112.5\ m$
0%
$62.5\ m$
0%
$32.5\ m$
0%
$50\ m$

Explanation

$u=10 m/s, a=5 m/s^2, t=5 s$

$S = ut+\dfrac{1}{2}at^2$

$S = 10\times 5+\dfrac{1}{2}\times 5\times 5^2$

$S = 112.5 m$

A toy car, initially moving with a uniform velocity of

$18 \ km h^{-1}$ comes to rest in

$2\ s$ . Find the retardation of the car in S.I units.

0%
$9 m s^{-2}$
0%
$25 m s^{-2}$
0%
$2.5 m s^{-2}$
0%
$5 m s^{-2}$

Explanation

Velocity,

$v = \dfrac{18\times1000 m}{60\times60 s} = 5 m/s$

Using formula

$v=u-at$ , where

$v=0$ and

$u=5 m/s$
Therefore, retardation,

$a = \dfrac{u}{t} = \dfrac{5 m/s}{2 s} = 2.5 m/s^2$

A body initially at rest starts moving with a constant acceleration

$2\ ms^{-2}$ . Calculate the distance traveled in 5 s.

0%
50 m
0%
12.5 m
0%
25 m
0%
5 m

Explanation

$S=ut+\dfrac{1}{2}at^2$

$S = 0\times 5+\dfrac{1}{2}\times 2\times 5^2 = 25 m$

Which of the following displacement-time graphs represents a uniform motion ?

A body moving with a constant acceleration travels the distances 3 m and 8 m respectively in 1 s and 2 s. Calculate the initial velocity.

0%
$4\, m\, s^{-1}$
0%
$0 \, m\, s^{-1}$
0%
$2\, m\, s^{-1}$
0%
$1\, m\, s^{-1}$

Explanation

$a=ut+\dfrac{1}{2}at^2$

Putting the values,

$3=u\times 1+\dfrac{1}{2}a\times 1^2$

$a=6-2u..............(i)$

Similarly,

$8=u\times 2+\dfrac{1}{2}a\times 2^2$

$a=4-u.................(ii)$

Equating (i) and (ii),

$6-2u=4-u$

$u = 2\ m/s$

Figure shows the displacement-time graph for the motion of two boys A and B along a straight road in the same direction. Which of the two has greater velocity?

0%
B
0%
A
0%
Both have same velocity
0%
None of the above

Explanation

From a graph of displacement vs time, the slope gives the rate of change of displacement or velocity

From the given graph the velocity of

$A$ is

$\dfrac{20}{4} = 5km/hr$

$\&$

Velocity of

$B$ is

$\dfrac{20}{2} = 10km/hr$

A train starts from rest and accelerates uniformly at a rate of

$2\ ms^{-2}$ for 10 s. It then maintains a constant speed for 200 s. The brakes are then applied and the train is uniformly retarded and comes to rest in 50 s. Find the maximum velocity reached.

0%
$2\, m\, s^{-1}$
0%
$10 \, m\, s^{-1}$
0%
$1\, m\, s^{-1}$
0%
$20\, m\, s^{-1}$

Explanation

Maximum velocity is attained at the end of uniformly accelerated motion.
From,

$v=u+at$
Maximum velocity,

$v=0+2\times 10 = 20\ m/s$

A train moving with a velocity of

$20\ ms^{-1}$ is brought to rest by applying the brakes in 5 s. Calculate the retardation.

0%
$4 m s^{-2}$
0%
$2 m s^{-2}$
0%
$1 m s^{-2}$
0%
$8 m s^{-2}$

Explanation

$u = 20 m/s, v=0, t=5 s$

$v=u+at$

$0=20+a\times 5$

$a = -4 m/s^2$
Therefore,

$\text{Retardation} =4 m/s^2$

A car travels a distance 100 m with a constant acceleration and average velocity of

$20\, ms^{-1}$ . The final velocity acquired by the car is

$25\, ms^{-1}$ . Find the acceleration of the car.

0%
$1\, m\, s^{-2}$
0%
$2\, m\, s^{-2}$
0%
$0\, m\, s^{-2}$
0%
$4\, m\, s^{-2}$

Explanation

$\text{Average velocity} = \dfrac{v+u}{2}$

$20 = \dfrac{25+u}{2}$

$u = 15\ m/s$

$v=u+at$

$25=15+a\times 5$
We get,

$a = 2\ m/s^2$

When brakes are applied to a bus, the retardation produced is

$25\ \ cm\ s^{-2}$ and the bus takes 20 s to stop. Calculate the initial velocity of the bus.

0%
$500 \, m\, s^{-1}$
0%
$5\, cm\, s^{-1}$
0%
$5\, m\, s^{-1}$
0%
$12.5 \, m\, s^{-1}$

Explanation

Retardation is

$25 cm/s^2$ , therefore, Acceleration,

$a=-0.25 m/s^2$

$t=20 s, v=0$
Putting the values in

$v=u+at$

$0 = u+(-0.25)(20)$

$u = 5 m/s$

Figure below shows the distance-time graph of three objects

$A, B$ and

$C$ . Study the graph and answer the following questions: which of the three is travelling the fastest?

0%
Car $C$
0%
Car $A$
0%
Car $B$
0%
Can not be determined

When a ball is thrown up vertically with velocity

${v}_{o}$ it reaches a maximum height of

$h$ . If one wishes to triple the maximum height then the ball should be thrown with velocity.

0%
$\displaystyle \sqrt{3}v_{0}$
0%
$\displaystyle 3v_{0}$
0%
$\displaystyle 9v_{0}$
0%
$\displaystyle \frac{3}{2}v_{0}$

Explanation

Maximum height means final velocity is zero.
Thus

$H=\dfrac{0-{u}^{2}}{-2g}=\dfrac{{u}^{2}}{2g}$
To triple maximum height, i.e. 3H, we have,

$3H=\dfrac{{u'}^{2}}{2g}$ or

${u'}^{2}=3{u}^{2}$ or,

$u'=\sqrt{3} u$

What does the odometer of an automobile measure?

0%
Speed
0%
Distance
0%
Acceleration
0%
All

An object is sliding down on an inclined plane. The velocity changes at a constant rate from 10 cm/s to 15 cm/s in 2 seconds. What is its acceleration?

0%
$5 cm/s^2$
0%
$7.5 cm/s^2$
0%
$2.5 cm/s^2$
0%
$12.5 cm/s^2$

Explanation

$a=\dfrac { v-u }{ t } =\dfrac {15-10}{2} =2.5\ cm/{ s }^{ 2 }$

A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of

$3.0\ ms^{-2}$ for

$8\ s$ . How far does the boat travel during this time?

0%
$96\ m$
0%
$192\ m$
0%
$12\ m$
0%
$72\ m$

Explanation

Given, initial velocity of boat is:

$u=0$
Acceleration is:

$a=3\ ms^{-2}$
Time is:

$t=8\ s$

Using second equation of motion:

$s=ut+\cfrac {1}{2}at^2$

$s=0\times 8+\cfrac {1}{2}\times 3\times 8^2=96\ m$ .

Which of the following relations is correct?

0%
Speed = $\frac { 1 }{ Distance\times Time }$
0%
Speed = $\frac { Time }{ Distance }$
0%
Speed = $\frac { Distance }{ Time }$
0%
Speed = Distance $\times$ Time

Explanation

Speed is defined as the distance travelled by an object per unit time.

$Speed = \dfrac{Distance}{time}$

Which of the following is NOT matched correctly?

0%
Speedometer: Speed
0%
Odometer : Odour
0%
Anemometer : Wind speed
0%
Stopwatch : Time

The table below shows the speed of a moving vehicle with respect to time.

Speed (m/s)	2	4	6	8	10
Time (s)	1	2	3	4	5

Find the acceleration of the vehicle.

0%
$2 m/s^2$
0%
$4 m/s^2$
0%
$6 m/s^2$
0%
$8 m/s^2$

Explanation

From the given data,

$v=2t$

Acceleration,

$a = \dfrac{v_2-v_1}{t_2-t_1}$

$= \dfrac{2(t_2-t_1)}{t_2-t_1}\ = 2\ m/s^2$

State whether given statement is True or False.
Both speed and velocity of a particle can be negative.

0%
True
0%
False

$2\ kg$ object has an initial velocity of

$u=10\ ms^{-1}$ . The retardation due to friction is

$2\ ms^{-2}$ . After what time does the object stop?

0%
$5\ s$
0%
$2\ s$
0%
$10\ s$
0%
$8\ s$

Explanation

Given:

Initial velocity,

$u = 10\ ms^{-1}$

Final velocity,

$v = 0\ ms^{-1}$

Acceleration,

$a = -2\ ms^{-2}$

Using first equation of motion:

$v=u+at$

$\Rightarrow 0 = 10 - 2t$

$\Rightarrow t=5\ s$

State whether given statement is True or False.
Is displacement independent of path?

0%
True
0%
False

With what speed should a car travel so that it can cover a distance of 5 km in 5 min?

0%
1 km/h
0%
5 km/h
0%
12 km/h
0%
60 km/h

Explanation

Given,

Distance cover

$d=5\ km$

Time

$t=5\ min$ or

$t=5/60\ hr$

Speed

$v=\dfrac{5}{5/60}=60\ km/hr$

Option D

A man runs

$10 \ km$ in

$30 \ min.$ What is his speed?

0%
$5 \ km/h$
0%
$10 \ km/h$
0%
$15 \ km/h$
0%
$20 \ km/h$

Explanation

We know that,

$Speed = \dfrac{distance}{time}$

Given:

Distance

$=10 \ km$

Time

$=30 min= \dfrac{30}{60} \ hr=0.5 \ hr$

Speed

$=\dfrac{10 \ km}{0.5 \ hr} = 20 \ km/h$

Option D is the answer.

Velocity is the (distance or displacement) per unit time.

0%
Displacement
0%
Both Displacement and Distance
0%
None
0%
Sum of both

Explanation

Velocity of an object is speed of that object plus direction in which object is moving.

It is given by

$Velocity=\dfrac{Displacement}{Time}$

Hence option A is correct.

The ratio of the heights from which two bodies are dropped is 3 : 5 respectively. The ratio of their final velocities is

0%
$\displaystyle \sqrt { 5 } :\sqrt { 3 }$
0%
$\displaystyle \sqrt { 3 } :\sqrt { 5 }$
0%
9 : 25
0%
5 : 3

Explanation

Initial velocity

$u = 0$

Using

$v^2-u^2 = 2gS$

$v^2 - 0 = 2gS$

$\implies \ v = \sqrt{2gS}$

$\implies \ v_1:v_2 = \sqrt{S_1}:\sqrt{S_2} = \sqrt{3}:\sqrt{5}$

From the displacement-time graph shown here, find the velocity of the body as it moves from B to C.

0%
$0\ ms^{-1}$
0%
$2\ ms^{-1}$
0%
$4\ ms^{-1}$
0%
$8\ ms^{-1}$

Explanation

We know that,

Area under the graph

Displacement during

$6^{th}$ sec

$d= 4\times 1 = 4m$

$Velocity = \dfrac {Displacement}{Time}$

$\Rightarrow v= \dfrac {4}{1}$

$\Rightarrow v= 4 ms^{-1}$

$\therefore$ Option (C) is correct.

2109308_269706_ans_2c374952428b484da6ae1fecf6e529d5.png

Graph : Distance vs Time

0%
Zero Speed
0%
Constant Speed
0%
Uniformly Increasing Speed
0%
Uniformly Decreasing Speed

The unit for the rate of change of velocity is

0%
$m s^{-2}$
0%
$m s^{-3}$
0%
$m s^{-1}$
0%
$m^{3} s^{-2}$

Explanation

The rate of change of velocity with respect to time is called acceleration.

GIven by

$a=\dfrac{change\quad in\quad velocity}{time}$

Putting units of velocity and time.

$a=\dfrac{m/s}{s}$ =

$m/s^2$ or

$ms^{-2}$ .Hence this is the unit of acceleration or rate of change of velocity with respect to time.

According to the following graph, what happens to the distance covered by the body from 0 -10 minutes?

0%
It goes on increasing
0%
It goes on decreasing
0%
It first increases and then decreases
0%
It first decreases and then increases

Explanation

We know that distance traveled by an object is the area under it speed time graph.

Now, in this case, as the area under the speed-time graph is increasing from 0-10 minutes. So, the distance will keep on increasing from 0-10 minutes.

Hence, correct answer is option

$A$

The odometer of a car reads

$57321.0\ km$ when the clock shows the time 8:30 AM. The distance moved by car if, at 8:50 AM, the odometer reading has changed to

$57336.0\ km$ and also the speed will be:

0%
$15\ km ,\ 45\, km/h$
0%
$10km ,$ $45\, km/h$
0%
$15km ,$ $55\, km/h$
0%
$20km ,$ $45\, km/h$

Explanation

Time from $8.30$ $AM$ to $8.50$ $AM$ = $20$ $min$

$20min=(20/60)h=(1/3)h$

(a) Distance covered = Second reading - First reading = $57336.0$ $km$ - $57321.0$ $km$ = $15$ $km$

(b) $Speed=\dfrac{Distance}{Time}$ = $\dfrac{15}{1/3}$ = $45km/h$

The rear view mirror of a car is a plane mirror. A driver is reversing his car at a speed of 2 m/s. The driver sees in his rear view mirror the image of a truck parked behind his car. The speed at which the image of the truck appears to approach the driver will be ________________.

0%
1 m/s
0%
2 m/s
0%
4 m/s
0%
8 m/s

CBSE Questions for Class 9 Physics Motion Quiz 6 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

CBSE Questions for Class 9 Physics Motion Quiz 6 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

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