the actual path travelled by the body between the initial and final positions

the shortest path between the initial and final positions

the longest possible path between the initial and final positions

MCQ Questions for Class 9 Physics Motion Quiz 8

A displacement-time graph of a body moving with uniform velocity is shown in the figure. Find out its velocity and its displacement at the end of 5 seconds.

0%
10 m/s, 125 m
0%
5 m/s, 225 m
0%
3m/s, 15 m
0%
15 m/s, 75 m

Explanation

From observing the graph, we say that displacement of the body at the end of 5 seconds is

$15 \ m$ .

Slope of the displacement-time graph gives the velocity of the body.

So, velocity of body

$v = slope = \dfrac{15-0}{5-0} = 3 \ m/s$

A ball is thrown vertically upwards from the top of a tower with a velocity of

$\displaystyle { 10\quad ms }^{ -1 }$ . The ball reaches the ground with the velocity

$\displaystyle { 30\quad ms }^{ -1 }$ . What is the height of the tower?

$\displaystyle (Take \quad g ={ 10\quad ms }^{ -2 })$

0%
40 m
0%
20 m
0%
60 m
0%
80 m

Explanation

Taking downward direction to be positive.

Initial velocity

$u = -10 \ m/s$

Final velocity

$v = 30\ m/s$

Let the height of the tower be

$H$ .

Acceleration

$a= g = 10 \ m/s^2$

Using

$v^2 - u^2 = 2aH$

$\therefore$

$30^2- (-10)^2 = 2\times 10 \ \times H$

$\implies\ H = 40 \ m$

A body whose speed in a particular direction is constant

0%
must be accelerating
0%
must be retarding
0%
has a constant velocity
0%
all the above

The final velocity of a body starting from rest and moving with an acceleration

$\displaystyle 2 \ { ms }^{ -2 }$ and covering a distance of 10 m is _____

$\displaystyle { ms }^{ -1 }$ .

0%
$\displaystyle \sqrt { 98 }$
0%
$\displaystyle \sqrt { 40 }$
0%
$\displaystyle \sqrt { 20 }$
0%
$\displaystyle \sqrt { 12 }$

Explanation

Initial velocity

$u = 0 \ m/s$
Let final velocity be

$v$ .
Acceleration

$a = 2 \ m/s^2$
Distance covered

$S = 10 \ m$
Using

$v^2 - u^2 = 2aS$

$\therefore$

$v^2 - 0 = 2\times 2\times 10$

$\implies \ v = \sqrt{40} \ m/s$

Two bodies of different masses

${m} _ {1}$ and

${m} _ {2}$ are dropped from two different heights

${h} _ {1}$ and

${h} _ {2}$ . What is the ratio of time taken by the two to reach the ground?

0%
$\displaystyle \sqrt { { h }_{ 1 } } :\sqrt { { h }_{ 2 } }$
0%
$\displaystyle \sqrt { { h }_{ 2 } } :\sqrt { { h }_{ 1 } }$
0%
$\displaystyle {m} _ {1} {h} _ {2} : {m} _ {2} {h} _ {1}$
0%
$\displaystyle {m} _ {1} {h} _ {1} : {m} _ {2} {h} _ {2}$

Explanation

Initial velocity of the stone

$u = o$

Acceleration

$a = g$

Let the height of the tower be

$h$ .

Using second equation of motion-

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

$\therefore$

$h = 0+\dfrac{1}{2}gt^2$

$\Rightarrow t = \sqrt{\dfrac{2h}{g}}$

We get,

$t\propto \sqrt{h}$

$\Rightarrow t_1:t_2 = \sqrt{h_1} : \sqrt{h_2}$

When brakes are applied the velocity of a car changes from

$\displaystyle { 30\ ms }^{ -1 } to\ { 10\ ms }^{ -1 }$ in 5 s. The acceleration produced in it is ____

$\displaystyle { ms }^{ -2 }$ .

0%
$4$
0%
$-4$
0%
$20$
0%
$-20$

Explanation

Initial speed

$u = 30 \ m/s$
Final speed

$v = 10 \ m/s$
Time

$t = 5 \ s$
Acceleration

$a = \dfrac{v-u}{t}$

$\implies \ a = \dfrac{10-30}{5} = -4 \ m/s^2$

With what speed should a bus travel so that it can cover a distance of 10 km in 5 min?

0%
60 km/hr
0%
10 km/hr
0%
12 km/hr
0%
120 km/hr

Explanation

$60$

$min$ =

$1hour$

$5$

$min$ =

$\dfrac { 1 }{ 60 } \times 5=\dfrac { 1 }{ 12 } h$

$Speed$ =

$\dfrac { Distance }{ Time }$

$Distance=10km$

$time=\dfrac{1}{12}h$

$speed$ =

$\dfrac { 10 }{ \dfrac { 1 }{ 12 } }$ =

$10\quad \times \quad 12$
=

$120$

$km/h$

A man runs

$20 \ km$ in

$30 \ min.$ What is his speed?

0%
$10 \ km/hr$
0%
$20 \ km/hr$
0%
$40 \ km/hr$
0%
$25 \ km/hr$

Explanation

Given:

Distance

$= 20 \ km$

Time

$=30 \ min =\dfrac{30}{60}hr= \dfrac { 1 }{ 2 } hr$

$\therefore$ Speed =

$\dfrac { Distance }{ time }$ =

$\dfrac { 20 \ km}{ \left( \dfrac { 1 }{ 2 }hr \right) }$

$\Rightarrow$ Speed

$=40 \ km/hr$

Option (C) is correct.

If a ball thrown vertically up attains a maximum height of

$80\ m,$ its initial speed is

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

0%
$\displaystyle { 40\ ms }^{ -1 }$
0%
$\displaystyle { 20\ ms }^{ -1 }$
0%
$\displaystyle { 50\ ms }^{ -1 }$
0%
$\displaystyle { 10\ ms }^{ -1 }$

Explanation

Maximum height attained

$H = 80 \ m$
Initial speed

$u = \sqrt{2gH}$

The final speed will be zero because at the highest point it changes its direction.

$\implies \ u = \sqrt{2\times 10\times 80} = 40 \ m/s$

The velocity of a retarding body changes from

$\displaystyle { 90\ ms }^{ -1 }$ to

$36\ ms^{ -1 }$ in

$9$ seconds. Find the retardation of a body?

0%
$\displaystyle { 20\ ms }^{ -2 }$
0%
$\displaystyle { 6\ ms }^{ -2 }$
0%
$\displaystyle { 18\ ms }^{ -2 }$
0%
$\displaystyle { 36\ ms }^{ -2 }$

Explanation

Initial velocity

$u = 90 \ ms^{-1}$

Final velocity

$v = 36 \ ms^{-1}$

Time,

$t = 9 \ s$

Using first equation of motion:

$v = u+at$

$\Rightarrow 36 = 90+a\times 9$

$\Rightarrow a = -6 \ m/s^2$

Thus retardation is

$6 \ m/s^2$

Which of the following formulae is correct?

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

Explanation

Speed of an object is defined as distance moved per unit time.It denotes how fast or how slow the object is moving.It is given by

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

A car is travelling at a certain speed. Its final velocity reaches 40 m/s in 10 seconds accelerating at the rate of 2.5 m/s

$\displaystyle ^{2}$ . Find the initial velocity of the car.

0%
20 m/s
0%
0 m/s
0%
15 m/s
0%
10 m/s

Explanation

Given,

Time

$(t) = 10$ seconds
Accelerartion

$(a) = 2.5$ m/s

$\displaystyle ^{2}$
We know

$v = u + at$

$\displaystyle \Rightarrow 40 m/s = u + 2.5\times 10 \ {m}/{s^{2}}$

$\displaystyle \Rightarrow 40 m/s = u + 25\ {m}/{s}$

$\Rightarrow u = (40 - 25)\ \ m/s$

$u = 15 m/s$

$\displaystyle \therefore$ initial velocity of the car

$= 15$ m/s

A car is travelling at 30 m/s. If its velocity increases to 50 m/s while accelerating at the rate of 2.5 m/s

$\displaystyle ^{2}$ find the time taken to reach that velocity.

0%
10 s
0%
8 s
0%
4 s
0%
5 s

Explanation

Given Final velocity (v) = 50 m/s
Initial velocity (u) = 30 m/s
Acceleration (a) = 2.5 m/s

$\displaystyle ^{2}$
We know v = u + at

$\displaystyle \therefore$ 50 = 30 + 2.5 t

$\displaystyle \Rightarrow$ 50 - 30 = 2.5 t

$\displaystyle \Rightarrow$

$\displaystyle \frac{20}{2.5}$ = t or 8 = t

$\displaystyle \therefore$ Time taken to accelerate = 8 seconds

Calculate the distance travelled by a man walking at a speed of 5 km/hr in 36 minutes.

0%
3 km
0%
4 km
0%
5 km
0%
6 km

Explanation

Given s = 5km/hr = 5

$\displaystyle \times \frac{5}{18}m/s$
t = 36 minutes = 36

$\displaystyle \times$ 60 sec

$\displaystyle \therefore$ distance travelled = s

$\displaystyle \times$ t
=

$\displaystyle \frac{25}{18}\times 36\times 60 = 3000 m$
=

$\displaystyle \frac{3000}{1000}km = 3 km$

When a satellite completes one revolution around the earth its displacement will be zero. True or false

0%
True
0%
False

The quantity which tells the distance an object travels in a certain time is called -

0%
acceleration
0%
speed
0%
velocity
0%
none of these

Explanation

The quantity which tells the distance an object travels in a certain time is called speed.

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

A stone dropped from the top of a tower reaches the ground with a velocity of 58.8 m/s in 6 seconds. Calculate the value of acceleration due to gravity.

0%
10 m/s $\displaystyle ^{2}$
0%
19.6 m/s $\displaystyle ^{2}$
0%
8.2 m/s $\displaystyle ^{2}$
0%
9.8 m/s $\displaystyle ^{2}$

Explanation

Here Initial velocity (u) = 0
Final velocity (v) = 58.8 m/s
Time (t) = 6 seconds
We know

$v = u +at$

$58.8 = a\times6$

$\displaystyle \Rightarrow 6 a = 58.8$

$\displaystyle \Rightarrow a = \frac{58.8}{6}$

$\displaystyle \therefore a = 9.8 m/s^{2}$

$\displaystyle \therefore$ The value of acceleration due to gravity = 9.8 m/s

$\displaystyle ^{2}$

A runner running along a circular track at a constant speed has _______ velocity.

0%
Variable
0%
Uniform
0%
Potential
0%
Parallel

If two bodies are moving with the same speed but in different directions they will have -

0%
same velocities
0%
same acceleration
0%
different velocities
0%
none of these

A boy cycles a distance of 640 m in 2 minutes and 40 seconds. What is the speed of the cycle?

0%
4 m/s
0%
8 m/s
0%
12 m/s
0%
16 m/s

Explanation

Distance covered = 640 m
Time taken = 2 min 40 sec

$2$

$\displaystyle \times 60 + 40 = 160s$ (because 1 minute =60seconds)

$\displaystyle \therefore Speed = \frac{Distance}{Time}= \frac{640}{160}=4m/s$

Can a moving object cover zero distance?

0%
Always
0%
Sometimes
0%
Never
0%
Can't be said from given condition

The velocity of a bullet is reduced from

$200\;\text{m/s}$ to

$100\;\text{m/s}$ while travelling through a wooden block of thickness

$10\;\text{cm}$ . Assuming it to be uniform, the retardation will be:

0%
$15\times10^4\;\text{m/s}^2$
0%
$10\times10^4\;\text{m/s}^2$
0%
$12\times10^4\;\text{m/s}^2$
0%
$14.5\;\text{m/s}^2$

Explanation

Using the formula, $\text{v}^2=\text{u}^2+2\text{as}$

$(100)^2=(200)^2-2\text{a}\times\displaystyle\frac{10}{100}$

$2\text{a}\times\displaystyle\frac{10}{100}=(200)^2-(100)^2=300\times100$

$\text{a}=\displaystyle\frac{3\times10^5}{2}=15\times10^4\;\text{m/s}^2$

The rate of change of velocity is called speed. True or false.

0%
True
0%
False

Can, a body in motion, have zero displacement?

0%
Yes, always
0%
Yes, provided its initial and final positions are same
0%
Never
0%
Yes, only if its initial and final positions are not same

A ball thrown vertically upwards with a velocity of 15 m/s reaches its highest point at 11.25 m in 1.5 sec. Find the total distance travelled by the ball and its position after 2 sec respectively. Take

$g=10 \ m/s^2$

0%
Total distance = 12 m, After 2 sec = 9.5 m.
0%
Total distance = 17.5 m, After 2 sec =7.5 m.
0%
Total distance = 14 m, After 2 sec = 9 m.
0%
Total distance = 12.50 m, After 2 sec = 10 m.

Explanation

The ball is thrown upwards.

$\therefore$ Acceleration due to gravity

$= -10 \ m/s^2$
Final velocity

$=\ 0m/s$
Initial velocity

$= 15\ m/s$
Distance

$= 11.25\ m$
Time

$= 1.5\ sec$
To find total distance traveled:

the distance traveled by the ball in upward and downward motion will be the sum of the height reached upto 1.5 sec plus height descended in next 0.5 sec.

Height descended,

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

$h=0+\dfrac{1}{2} (-10)(0.5)^2$

$h=1.25 m$

Hence, the total distance will be

$11.25\ m + 1.25\ m = 12.50\ m$

To find position after

$2\ sec$
Using the formula:

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

$s=15\times 2+\dfrac{1}{2}\times -10 \times (2)^2$

$s=30+(-40)$

$s=-10 m$

So the option D is correct.

Speed can be positive, zero or negative. True or false

0%
True
0%
False

Velocity is the distance travelled by a moving body per unit time. True or false.

0%
True
0%
False

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

$25 cm{s}^{-2}$ and the bus takes

$20s$ to stop. Calculate the initial velocity of the bus.

0%
$5 m{s}^{-1}$
0%
$10 m{s}^{-1}$
0%
$15 m{s}^{-1}$
0%
$20 m{s}^{-1}$

Explanation

Final Velocity,

$v=0$

Retardation,

$a=25\ cm s^{-2}=0.25\ ms^{-2}$

Time,

$t=20s$

Let, initial velocity

$=u$
According to newton first law

$v=u-at$

$0=u-(0.25\times 20)$

$u=5\ ms^{-1}$

A body, initially at rest, starts moving with a constant acceleration

$2 m{s}^{-2}$ . Calculate the distance travelled in 5 s.

0%
$10m$
0%
$15m$
0%
$25m$
0%
$50m$

Explanation

Initial velocity,

$u=0$

Acceleration,

$a=2ms^{-2}$

Time,

$t=5s$

Distance traveled,

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

$s=(0\times 5)+(\dfrac{1}{2}\times 2\times 5^2)=25m$

A car was moving at

$1 m{s}^{-1}$ when brakes were applied. The brakes offered a retardation of

$0.5 m{s}^{-2}$ . If the brakes are applied for

$2 s$ , after

$2 s$

0%
the car moves at a reduced speed of $0.5 m{s}^{-1}$
0%
the car moves at a reduced speed of $0.25 m{s}^{-1}$
0%
the car stops
0%
the car moves with an increased speed of $2 m{s}^{-1}$

Explanation

Initial velocity,

$u = 1 m{s}^{-1}$
Retardation,

$a = 0.5 m{s}^{-2}$
Time,

$t = 2 s$
Final velocity

$= v$
Using equation of motion,

$a = \dfrac{u-v}{t}$

$0.5 = \dfrac{1 - v}{2}$

$v = 1 - 1 = 0$
So, final velocity is

$0$ .

After

$2 s$ , the body stops.

Find the distance which the car travels before the brakes are applied.

0%
$75m$
0%
$100m$
0%
$120m$
0%
$125m$

Explanation

Before brakes are applied, the car was moving with uniforms velocity.

Distance = uniform velocity

$\times$ time

$=(25\times 5)m=125m$

Find the acceleration of the vehicle.

0%
$0.1\ A m{s}^{-2}$
0%
$0.3\ m{s}^{-2}$
0%
$0.5\ m{s}^{-2}$
0%
$1\ m{s}^{-2}$

Explanation

Given,

Initial velocity

$u=30\ kmph=\dfrac{30\times 1000}{60\times 60}ms^{-1}=\dfrac{25}{3}ms^{-1}$

After 2 sec, its velocity

$v=33.6\ kmph=\dfrac{33.6\times 1000}{60\times 60}ms^{-1} = \dfrac{28}{3}ms^{-1}$

$t_1= 2s$

$\therefore a=\dfrac{v-u}{t_1}=\dfrac{\frac{28}{3}-\frac{25}{3}}{2}ms^{-2}=0.5ms^{-2}$

A train is moving with a velocity of

$90\ kmph$ . When brakes are applied, the retardation produced is

$0.5\ m{s}^{-2}$ . Find the velocity after

$10 s$ .

0%
$20 m{s}^{-1}$
0%
$25 m{s}^{-1}$
0%
$30 m{s}^{-1}$
0%
$40 m{s}^{-1}$

Explanation

Initial Velocity,

$u=90\ kmph=\dfrac{90\times 1000}{60\times 60}ms^{-1}=25\ ms^{-1}$

Retardation,

$a=-0.5\ ms^{-2}$

Time,

$t=10\ s$

Using equation of motion

$v=u+at$

$v=25-(0.5\times 10)=20\ ms^{-1}$

A train is moving with a velocity of

$90\ kmph$ . When brakes are applied, the retardation produced is

$0.5 m{s}^{-2}$ . Find the time taken by the train to come to rest.

0%
$20s$
0%
$30s$
0%
$40s$
0%
$50s$

Explanation

Initial Velocity,

$u=90\ kmph=25\ ms^{-1}$

Retardation,

$a=-0.5\ ms^{-2}$

Final velocity,

$v=0$

Let time

$= t$

$v=u+at$

$0=25-(0.5\times t)$

$t=50\ s$

A car travelling at

$2 m{s}^{-1}$ undergoes an acceleration of

$1 m{s}^{-2}$ . How long will it take to double its velocity?

0%
$1s$
0%
$2s$
0%
$3s$
0%
$4s$

Explanation

$u=2\ ms^{-1}$

$a=1\ ms^{-2}$

$v=2u=4\ ms^{-1}$

$\therefore t=\dfrac{v-u}{a}=\dfrac{4-2}{1}s=2\ s$

A car moving at

$2.5 m{s}^{-1}$ doubles its velocity with an acceleration of

$0.5 m{s}^{-2}$ in some time. If the same car travels at

$1.5 m{s}^{-1}$ , what will be its final velocity if same acceleration acts on it for same time?

0%
$2 m{s}^{-1}$
0%
$3 m{s}^{-1}$
0%
$4 m{s}^{-1}$
0%
$5 m{s}^{-1}$

Explanation

$u=2.5\ ms^{-1}$

$v=2u=5\ ms^{-1}$

$a=0.5ms^{-2}$
Let us use:

$v=u+at$
or,

$5=2.5+(0.5\times t)$
or,

$t=\dfrac{2.5}{0.5}=5\ s$
So, using this value, but different initial velocity

$u$ =

$1.5ms^{-1}$ we get:

$v=u+at=1.5+(0.5\times 5)=4ms^{-1}$

Distance is

0%
the shortest path between the initial and final positions
0%
the longest possible path between the initial and final positions
0%
the actual path travelled by the body between the initial and final positions
0%
none of these

When a body moves in an uniform circular motion, changes in velocity and speed will be

0%
velocity changes but speed remains constant
0%
velocity remains constant but speed changes
0%
both remains constant
0%
both changes

The odometer of a car reads

$57321.0 km$ when the clock shows the time

$08:30$ AM. What is the distance moved by the car, if at

$08:50 AM$ , the odometer reading has changed to

$57336.0 km$ ?

0%
$75\ km$
0%
$50\ km$
0%
$15\ km$
0%
$10\ km$

Explanation

An odometer is a device that gives us the distance covered by a car.

Distance covered by the car in the given time is

$d = 57336.0 - 57321 = 15.0 \ km$

Find the total distance travelled.

0%
$460 \ m$
0%
$4600 \ m$
0%
$46 \ km$
0%
$460 \ km$

Explanation

Given, initially the train accelerates at a rate

$a_1$ for time

$t_1$

$u_1=0$

$a_1=2 \ ms^{-2}$

$t_1=10 \ s$

Using the second equation of motion , displacement is given by,

$s_1=u_1t_1+\dfrac{1}{2}a_1t_1^2=(0\times 10)+(\dfrac{1}{2}\times 2\times 10^2)=100 \ m$

Calculating the velocity

$v_1$ of the train after time

$t_1$

Using the first equation of motion, to calculate the final velocity,

$v_1=u_1+a_1t_1= 0+2 \times 10 = 20 \ ms^{-1}$

Now, the train moves at constant speed

$v_1$ for time

$t_2=200s$

Hence, displacement

$s_2$ is given by

$\therefore s_2=v_1\times t_2=(20\times 200)=4000 \ m$

Now, brakes are applied so that train comes to rest in time

$t_3$
final velocity

$w=0$
time taken

$t_3=50 \ s$

acceleration is given by,

$\therefore a_2=\dfrac{w-v_1}{t_3}=-0.4 \ ms^{-2}$

Using the second equation of motion, to calculate the distance travelled,

$s_3=(v_1\times t_3)+(\dfrac{1}{2}\times a_2\times t_3^2)=(20\times 50) -(\dfrac{1}{2}\times 0.4\times 50^2)$

$s_3=500 / m$

Hence, total distance ,

$s=s_1+s_2+s_3=(100+4000+500) \ m=4600 \ m$

A ball is initially moving with a velocity

$0.5 m{s}^{-1}$ . Its velocity decreases at a rate of

$0.05 m{s}^{-2}$ . How much time will it take to stop?

0%
$5s$
0%
$10s$
0%
$15s$
0%
$20s$

Explanation

Given,

$u=0.5\ ms^{-1}$

$a=-0.05\ ms^{-2}$ and

$v=0$

Now,

$v=u+at$

$0=0.5+(-0.05\times t)$

$t=10\ s$

A particle starts to move in a straight line from a point with velocity

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

$- 2.0 m{s}^{-2}$ . Find the position of the particle at

$t = 5 s$

0%
$25m$
0%
$-25m$
0%
$50m$
0%
$-50m$

Explanation

Initial Velocity,

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

Acceleration,

$a=-2\ ms^{-1}$

Time,

$t=5\ s$

Displacement,

$s$

Using Newton's second equation of motion,

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

$s=(10\times 5)-(\dfrac{1}{2}\times 2\times 5^2)$

$s=25\ m$

Larger the slope of a displacement-time graph

0%
lesser the velocity
0%
higher the velocity
0%
lesser the acceleration
0%
higher the acceleration

The slope of displacement-time graph being negative implies

0%
that the object moves towards the reference point
0%
that the object moves away from the reference point
0%
that the object is stationary
0%
none of these

Velocity, being a vector quantity, positive velocity means

0%
that the body is moving away from the reference point
0%
that the body is moving towards the reference point
0%
that the body is stationary
0%
nothing specifically

A bullet emerge from a barrel of length 1.2 m with a speed of

$\displaystyle 640{ ms }^{ -1 }$ . Assuming constant acceleration, the approximate time that it spends in the barrel after the gun is fired is

0%
4 ms
0%
40 ms
0%
400 $\displaystyle \mu s$
0%
1 s

Explanation

Given that,

$\displaystyle s=1.2m$ and

$\displaystyle v=640{ ms }^{ -1 }a=?;u=0;t=?$
We have the third equation of motion

$\displaystyle 2as={ v }^{ 2 }-{ u }^{ 2 }$

$\displaystyle 2a\times 1.2=640\times 640$

$\Rightarrow \displaystyle a=\frac { 640 \times 640 }{ 2 \times 1.2 }$

$\Rightarrow \displaystyle a=\frac { 512 \times 10^3 }{ 3 }$
and by first equation of motion

$\displaystyle v=u+at$
or

$\displaystyle t=\frac { v }{ a } = \frac{640 \times 3}{512} \times 10^{-3} =\frac { 15 }{ 4 } \times { 10 }^{ -3 }=3.75\times { 10 }^{ -3 }s\approx 4ms$

If a displacement-time graph of an object is a straight line parallel to the time axis, the object is

0%
stationary
0%
moving away from the reference point
0%
moving towards the reference point
0%
moving with an infinite velocity

A vehicle moving with a constant acceleration from

$A$ to

$B$ in a straight line

$AB$ , has velocities

$u$ and

$v$ at

$A$ and

$B$ respectively.

$C$ is the mid point of

$AB$ . If time taken to travel from

$A$ to

$C$ is twice the time to travel from

$C$ to

$B$ then the velocity of the vehicle

$v$ at

$B$ is:

0%
$5u$
0%
$6u$
0%
$7u$
0%
$8u$

Explanation

Let the velocity at A be

$u$ and acceleration be

$a$ .

Thus . using the third equation of motion ,

$v^2=u^2+2as$

where

$s$ is the distance between A and B.

From the first equation of motion,

Time taken to travel from A to C is

$\dfrac{v'-u}{a}$ , where

$v'$ is the velocity of vehicle at

$C$ .

Similarly time taken to travel from C to B is

$\dfrac{v-v'}{a}$

Thus

$\dfrac{v'-u}{a}=2\dfrac{v-v'}{a}$

$\implies v'=\dfrac{u+2v}{3}$

Also

$v'^2-u^2=2a\dfrac{s}{2}=as$

Thus

$v^2=u^2+2(v'^2-u^2)=2v'^2-u^2$

$=2(\dfrac{u+2v}{3})^2-u^2$

On simplification, we get,

$v^2-8uv+7u^2=0$

$v^2-uv-7uv+7u^2=0$

$(v-u)(v-7u)=0$

$v=u,7u$

But , since the body is moving with constant acceleration ,

$v>u$

$\implies v=7u$

A car accelerates uniformly so that it goes from a velocity of

$20 {m}/{s}$ to a velocity of

$40 {m}/{s}$ in

$5$ seconds. Find out the acceleration of a car?

0%
$0.2 {m}/{{s}^{2}}$
0%
$4 {m}/{{s}^{2}}$
0%
$5 {m}/{{s}^{2}}$
0%
$10 {m}/{{s}^{2}}$
0%
$80 {m}/{{s}^{2}}$

Explanation

Given :

$u = 20$ m/s

$v = 40$ m/s

$t =5$ s

Using

$v = u+ at$

$\therefore$

$40 = 20 + a (5)$

$\implies a = 4$

$m/s^2$

A stone dropped from the top of the tower reaches ground in

$4\ sec$ . Height of the tower is

$(g = 10m/s^{2})$

0%
$20m$
0%
$40m$
0%
$60m$
0%
$80m$

Explanation

Initial speed of the stone

$u =0 m/s$

Time taken by the stone to reach the ground

$t = 4 s$

Height of the tower

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

$\therefore$

$H = 0 +\dfrac{1}{2}\times 10\times 4^2$

$\implies$

$H = 80 m$

CBSE Questions for Class 9 Physics Motion Quiz 8 - MCQExams.com

0:0:1

Practice Class 9 Physics Quiz Questions and Answers

CBSE Questions for Class 9 Physics Motion Quiz 8 - MCQExams.com

0:0:1

Practice Class 9 Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.