MCQ Questions for Class 9 Physics Motion Quiz 11

A body is dropped from a height

$39.2\ m$ . After it crosses the half distance, the acceleration due to gravity ceases to act. Then the body will hit the ground with a velocity of (Take

$g=9.8\ ms^{-2}$ )

0%
$19.6\ ms^{-1}$
0%
$20\ ms^{-1}$
0%
$1.96\ ms^{-1}$
0%
$196\ ms^{-1}$

Explanation

Given that,

Height $h = 39.2\ m$

Initial velocity $u = 0$

Now, from equation of motion

After crosses half distance, the acceleration due to gravity ceases to act

So,

$h=\dfrac{39.2}{2}$

$h=19.6\,m$

${{v}^{2}}-{{u}^{2}}=2gh$

${{v}^{2}}=2\times 9.8\times 19.6$

$v=19.6\,m/s$

Hence, the velocity is $19.6\ m/s$

A body travels a distance x in first two second and distance y in next two second. The relation between

$x$ and

$y$ is

0%
$y=4x$
0%
$y=x$
0%
$y=3x$
0%
$y=2x$

Explanation

$\rightarrow$ Distance Travelled in

$2\ sec$

$S= ut + \dfrac{1}{2} at^{2}$ u=0

t=2$$

$\therefore x=0+ \dfrac{1}{2} a2^{2} = x= 2a \Rightarrow a = \dfrac{x}{2}$

Distance travelled in

$4\ sec$ Distance travelled in

$2\ sec =$ Distance travelled in next

$2\ sec$

$\therefore y =\dfrac{a}{2} (4)^{2}- \dfrac{a}{2} (2)^{2}= 8a-2 a = 6a$

$\therefore y= 6a \Rightarrow y=6 \dfrac{(x)}{2}$

$\left[ \because x=2a \ a=\dfrac{x}{2} \right]$

$\therefore y= 3x$

Answer is

$C$

A stone is thrown horizontally with the velocity

$v=15\ m/s$ . from a tower of height

$h=25\ m$ .

0%
The time during with the stone in motion is $2.25$
0%
The distance from tower base to the ground is 33.75
0%
The velocity $v$ with which it will touch the ground will be $20.66\ m/s$
0%
The angle $\theta$ , the trajectory of the stone makes with horizontal at the point stone touches the ground ( $50.7^{o}$ )

Explanation

Given,

Vertical velocity is zero

Horizontal velocity is $v=15\,m/s$

Height, $h=25\,m$

Time period $t=\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2\times 25}{9.81}}=2.25\ \sec$

Range, $R=vt=15\times 2.25=33.75\,m$

The speed-time graph of a particle moving along a solid cover is shown below. The distance traversed by the particle from

$t = 0$ to

$t = 3$ is:-

0%
$\dfrac{9}{2} m$
0%
$\dfrac{9}{4} m$
0%
$\dfrac{10}{3} m$
0%
$\dfrac{10}{5} m$

Explanation

Distance covered is equal to the area under the graph in given interval of time.

$\Rightarrow Distance = area = \dfrac{1}{2} \times 3\times 1.5 = \dfrac{9}{4}m$

Thereefore, B is correct option.

In uniform circular motion the

0%
acceleration is parallel (or antiparallel) to the velocity
0%
acceleration is perpendicular to the velocity
0%
acceleration is vertical, while the velocity can be any direction
0%
acceleration is vertical and the velocity is horizontal

Ball

$A$ is thrown up vertically with speed

$10 \ m/s$ . At the same instant another ball

$B$ is released from rest at height

$h$ . At time

$t$ , the speed of

$A$ relative to

$B$ is

0%
$10 \ m/s$
0%
$10 - 2 \ gt$
0%
$\sqrt{10^2 - 2gh}$
0%
$10 - gt$

Explanation

$u_A=10 ,\ u_B=0\ ,\ a_A=-g\ , a_B=-g$

$u_{AB}=10\ , \ a_{AB}=0$

$v_{AB}=u_{AB}+a_{AB}t$

$v_{AB}=10m/s$

A car covers a distance of

$2 \ km$ in

$2.5$ minutes. If it covers half of the distance with speed

$40 \ km/hr$ , the rest distance it shall cover with a speed of:

0%
$56 \ km/hr$
0%
$60 \ km/hr$
0%
$48 \ km/hr$
0%
$50 \ km/hr$

Explanation

Given that,

Time $t=2.5\min$

Distance $d=2km$

Now, firstly for half distance

$v=\dfrac{d}{t}$

$t=\dfrac{d}{v}$

$t=\dfrac{60\times 60}{40}$

$t=90\,s$

Now, total time is

$t=150-90$

$t=60\,s$

Now, the speed is

$v=\dfrac{d}{t}$

$v=\dfrac{1\times 60}{1}$

$v=60\,km/h$

Hence, the speed is $60 km/h$

A ball is projected vertically upwards. Its speed at half of maximum height is

$20 \ m/s$ . The maximum height attained by it is [Take

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

0%
$35 \ m$
0%
$15 \ m$
0%
$25 \ m$
0%
$40 \ m$

Explanation

$(D)$

using :

$v^2-u^2=2as$

at max height:

$v=0 , \Rightarrow u=\sqrt{2gh_{max}}$

given:

$s=\dfrac{h_{max}}{2} , v=20$

using :

$v^2-u^2=2as$

$20^2-(\sqrt{2gh_{max}})^2=-2g(\dfrac{h_{max}}{2})$

$400=gh_{max}$

$h_{max}=40$

A ball is thrown up under gravity

$(g=10 m/sec^2)$ . Find its velocity after 1.0 sec at a height of 10 m

0%
$5 m/sec^2$
0%
$5 m/sec$
0%
$10 m/sec$
0%
$15 m/sec$

Explanation

suppose velocity of projection

$=u$

then velocity at height

$h$ will be

$v=\sqrt { { u }^{ 2 }-2gh }$

putting

$h=10m,g=10m/{s}^{2}$

we get

$v=\sqrt { { u }^{ 2 }-200 } ...(1)$

but at

$t=1$ second is the time taken to reduce

$u$ to

$v$

$v=u-g(1)=u-g$

$u=v+g$

putting this in equation 1

${v}^{2}={u}^{2}-200={(v+g)}^{2}-200$

${v}^{2}={v}^{2}+{g}^{2}+2vg-200$

$\Rightarrow$

$0=100+2v(10)-200$

$0=-100+20v$

$v=5m/s$

The ratio of SI units to CGS units of retardation is.

0%
$10^2$
0%
$10$
0%
$10^3$
0%
$10^{-2}$

A particle travels half of the distance of a straight journey with a speed

$6 \ m/s$ . The remaining part of the distance is covered with speed

$2 \ m/s$ for half of the time and with speed

$4 m/s$ for other half of time. The average speed of the particle is :

0%
$3 \ m/s$
0%
$4 \ m/s$
0%
$\dfrac{3}{4} \ m/s$
0%
$5 \ m/s$

Explanation

Let total distance covered in the entire journey is $2x$ .

Let $t_1$ is the time to cover half of the distance with a speed of $6 \ m/s$ .

So, ${{t}_{1}}=\dfrac{x}{6}............(1)$

Let, for the second part total time is $t_2$ .

Let the distance covered in first half time with a speed of $2 \ m/s$ is $x_1$ . So, ${{x}_{1}}=\dfrac{{{t}_{2}}}{2}\times 2={{t}_{2}}............(2)$

For the second half the distance covered with a speed of $4 \ m/s$ is $x_2$ . So, ${{x}_{2}}=\dfrac{{{t}_{2}}}{2}\times 4=2{{t}_{2}}..........(3)$

Now,

${{x}_{1}}+{{x}_{2}}=x$

${{t}_{2}}+2{{t}_{2}}=x$

${{t}_{2}}=\dfrac{x}{3}............(3)$

Hence, $total \ time = t_1+t_2 = \dfrac{x}{6} + \dfrac{x}{3}$

Now , $average\,\,speed=\dfrac{total\,\,distance}{total\,\,time}$

$s=\dfrac{2x}{\dfrac{x}{6}+\dfrac{x}{3}}=\dfrac{2x}{3x}\times 6$

$s=4\,m/s$

In a uniform circular motion -

0%
both velocity and acceleration are constant.
0%
acceleration and speed are constant but velocity changes.
0%
both acceleration and velocity change
0%
both acceleration and speed are constant

A particle starts from rest with acceleration

$2\ m/s^2$ , The distance moved by the particle in 5 sec is:

0%
22 m
0%
28 m
0%
25 m
0%
13 m

Explanation

Given:

Acceleration $a=2\,m/{{s}^{2}}$

Time $t=5\,s$

Initial velocity $u=0$

We know from the equation of motion,

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

$\Rightarrow s=0+\dfrac{1}{2}\times 2\times 5\times 5$

$s=25\,m$

Hence, the distance moved by the particle is $25\ m$

A train starting from a railway station and moving with uniform acceleration attains a speed

$40km\ h^{-1}$ in

$10$ minutes. Find its acceleration.

0%
$0.0185 m/s^2$
0%
$0.0195 m/s^2$
0%
$0.0165 m/s^2$
0%
$0.0135 m/s^2$

Explanation

Here we have, initial velocity,

$v=40$ km/h

$=11.11$ m/s

Time

$t=10$ minutes

$=60\times 10=600$ secs

We know that

$v=u+at$

$\Rightarrow 40\ km{hr}^{-1}=0+a\times 10$ m

$\Rightarrow 11.11 \ m{s}^{-1}=a\times 600$ secs

$\Rightarrow \dfrac{11.11}{600}=0.0185\ m{s}^{-2}$

A lift is moveing with a uniform downward acceleration of

$2 m/s^2$ . A ball is dropped from height 2 cm from the floor of lift. Find the time take after which ball will strike the floor?

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

0%
$\sqrt { 2 } sec$
0%
2 sec
0%
$\frac { 1 }{ 2 } sec$
0%
$\frac { 1 }{ \sqrt { 2 } } sec$

Explanation

Given that,

Acceleration of lift $a=2\,m/{{s}^{2}}$

We know that,

Acceleration of ball relative to lift = acceleration of ball relative to ground – acceleration of lift with respect to ground

$a'=-10\hat{j}-\left( -2\hat{j} \right)$

$a'=8\hat{j}$

Now, from equation of motion

$s=ut-\dfrac{1}{2}a{{t}^{2}}$

$s=0+\dfrac{1}{2}\times 8\times {{t}^{2}}$

${{t}^{2}}=\dfrac{1}{2}$

$t=\dfrac{1}{\sqrt{2}}\,s$

Hence, the is $\dfrac{1}{\sqrt{2}}\,s$

A train accelerates from rest at a constant rate

$\alpha$ for distance

$x_1$ and time

$t_1$ . After that it retards to rest at constant rate

$\beta$ for distance

$x_2$ and time

$t_2$ . Which of the following relations is correct?

0%
$\dfrac{x_1}{x_2} = \dfrac{\alpha}{\beta} = \dfrac{t_1}{t_2}$
0%
$\dfrac{x_1}{x_2} = \dfrac{\beta}{\alpha} = \dfrac{t_1}{t_2}$
0%
$\dfrac{x_1}{x_2} = \dfrac{\alpha}{\beta} = \dfrac{t_2}{t_1}$
0%
$\dfrac{x_1}{x_2} = \dfrac{\beta}{\alpha} = \dfrac{t_2}{t_1}$

Explanation

According to question.................

$\begin{array}{l} The\, law\, of\, \, kinematics\, for\, constant\, \, acceleration\, \, and\, \, retardation, \\ to\, find\, \, the\, \, relationship. \\ For\, the\, \, 1st\, half\, of\, accelerated\, \, motion,\, \, we\, \, can\, \, write, \\ { v^{ 2 } }=2\alpha { x_{ 1 } }......................(i)\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (v\, \, is\, \, the\, \, velocity\, \, \, gained\, \, \, \, after\, \, \, time\, \, { t_{ a } }) \\ Now,at\, the\, \, \, end\, \, of\, first\, \, { t_{ 1 } }s,\, \, if\, it\, \, gains\, velocity\, \, v,then, \\ v=0+\alpha { t_{ 1 } } \\ So,putting\, \, the\, \, value\, \, \, of\, \, v\, \, in\, \, equation\, \, \, (i)\, \\ we\, \, get,{ \alpha ^{ 2 } }{ ({ t_{ 1 } })^{ 2 } }=2\alpha { x_{ 1 } } \\ or,{ x_{ 1 } }=\alpha \frac { { { { ({ t_{ 1 } }) }^{ 2 } } } }{ 2 } \\ So,for\, \, the\, \, \, rest\, \, \, half\, \, \, of\, \, \, the\, \, journey,\, \, wecan\, say, \\ { 0^{ 2 } }={ (\alpha { t_{ 1 } })^{ 2 } }-2\beta { x_{ 2 } }\, \, or,{ x_{ 2 } }={ \alpha ^{ 2 } }\frac { { { { ({ t_{ 1 } }) }^{ 2 } } } }{ { 2\beta } } \\ And,0=\alpha { t_{ 1 } }-\beta { t_{ 2 } } \\ So,\frac { \alpha }{ \beta } =\frac { { { t_{ 2 } } } }{ { { t_{ 1 } } } } \\ And,\frac { { { x_{ 1 } } } }{ { { x_{ 2 } } } } =\frac { { \frac { { \alpha { { ({ t_{ 1 } }) }^{ 2 } } } }{ 2 } } }{ { { \alpha ^{ 2 } }\frac { { { { ({ t_{ 2 } }) }^{ 2 } } } }{ { 2\beta } } } } =\frac { \beta }{ \alpha } \\ So,we\, \, can\, \, conclude,\frac { { { x_{ 1 } } } }{ { { x_{ 2 } } } } =\frac { \beta }{ \alpha } =\frac { { { t_{ 1 } } } }{ { { t_{ 2 } } } } \end{array}$

so the option is B.

A car, starting from rest, accelerates at the rate

$f$ through a distance

$S$ , then continues at constant speed for time

$t$ and then decelerates at the rate

$\frac{f}{2}$ to come to rest. If the total distance traversed is

$15S$ , then

0%
$S=\frac{1}{2}\ ft^{2}$
0%
$S=\frac{1}{4}\ ft^{2}$
0%
$S=\frac{1}{72}\ ft^{2}$
0%
$S=\frac{1}{6}\ ft^{2}$

Explanation

Let the value of car after covering

$S$ distance by

$x$ . Then

By third equation of motion

$v^2-(0)^2=2fs\ \Rightarrow \ v=\sqrt {2fs}$

constant speed

$v$ be

$8$ , then

$s'=v.t\ \Rightarrow \ s=(\sqrt {2fs})t$

While rerefraction, let the distance convered be

$s'$ . By

By third equation of motion

$(0)^2-v^2=(2)\left (\dfrac {-f}{2}\right)s"$

$\Rightarrow \ s"=\dfrac {2fs}{f}$

$\Rightarrow \ s"=2s$

So, Total distance

$(15s)=s+s'+s"$

$\Rightarrow \ 15s=s+\sqrt {2fs}t+t2s$

$\Rightarrow \ (12)^2s^2=2f+t^2s\ \Rightarrow \ s=\dfrac {ft^2}{72}$

OPtion

$C$ is correct.

A stone falls from a balloon that is descending at a uniform rate of

$12\, ms^{-1}$ . The displacement of the stone from the point of release after 10 sec is :-

0%
$490\ m$
0%
$610\ m$
0%
$510\ m$
0%
$725\ m$

Explanation

Given,

$u=12m/s$ , the velocity of the stone is same as ballon

$t=10\ sec$

$g=9.8\ ms^{-2}$

From 2nd equation of motion,

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

$\Rightarrow s=12\times 10+\dfrac{1}{2}\times 9.8\times 10\times 10$

$\Rightarrow s=120+490$

$\Rightarrow s=610m$

The correct option is B.

$2nd$ equation of motion

$(s=ut + \dfrac{1}{2}at^2)$ can be derived graphically form

$v-t$ graph by finding

0%
Slope
0%
Area
0%
Either(A) or (B)
0%
Insufficient data

Explanation

$2nd$ equation of the motion

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

From

$v-t$ graph-

The area under the

$v-t$ graph is the displacement of the particle.

The slope represents the acceleration of the particle.

1221523_1208058_ans_1dd6a4d22d274debb30680afe071ab67.png

A body starting from rest moving with uniform acceleration has a displacement of

$16 \ m$ in first

$4 \ s$ and

$9 \ m$ in first

$3 \ s$ . The acceleration of the body is

0%
$1 \ ms^{-1}$
0%
$2 \ ms^{-1}$
0%
$3 \ ms^{-1}$
0%
$4 \ ms^{-1}$

Calculate the speed of the body as if moves from: B to C.

0%
$3\ cm/s$
0%
$1.5\ cm/s$
0%
$Zero$
0%
$-3\ cm/s$

Explanation

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

At point B, the distance at 5 sec = 3 cm and at point C, the distance at 7 sec = 3cm

Change in distance = (3-3) cm = 0

Hence speed is zero

2063148_1210590_ans_0a377128228b42b6a88a4631171537da.png

A helicopter is flying at a height of

$500m$ . If all of sudden its engines stop working, the helicopter will fall on the earth in __ seconds?

0%
$10$
0%
$12$
0%
$15$
0%
$20$

Explanation

Initial velocity,

$u=0$ ; only downward velocity considered

acceleration

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

Displacement,

$s=500$ m

From Newton's second equation of motion,

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

$\Rightarrow 500=\dfrac{1}{2}\times 10 \times t^2$

$\Rightarrow t^2=100$

$\Rightarrow t=10s$

The displacement - time graph of a particle moving in a straight line is shown in the figure. Then

0%
Velocity at point $D$ is greater than that at point $B$
0%
Speed of the particle is increasing continuously
0%
Acceleration is uniform
0%
The particle is first uniformly accelerated and then retarded

Explanation

The slope of the displacement - time graph at a point gives us the velocity at that point. From the graph, it is clear that the slope at point

$D$ is greater than that at point

$B$ . Hence the velocity at point

$D$ is greater than that at point

$B$ . So, option (A) is correct.

A man is

$45\ m$ behind the bus when the bus starts acceleration from rest with acceleration

$2.5\ ms^{-2}$ . With what minimum velocity should the man start running to reach the bus?

0%
$12\ ms^{-1}$
0%
$14\ ms^{-1}$
0%
$15\ ms^{-1}$
0%
$16\ ms^{-1}$

A particle is released from rest from a tower of height 3h.The ratio of times to fall equal height h,i.e., t

$_{1}$ : t

$_{2}$ :t

$_{3}$ is

0%
$\sqrt3 :\sqrt2$ : 1
0%
3 : 2 : 1
0%
9 :4 : 1
0%
1 : ( $\sqrt2$ - 1) :( $\sqrt3 - \sqrt2 -1$ )

Explanation

Time in falling first height

$h$ will be

$t_1=T_1=\sqrt[2]{\dfrac{2h}{g}}$

Time in falling first height

$2h$ will be

$T_2=\sqrt[2]{\dfrac{2(2h)}{g}}$

so the time in falling second

$h$ will be

$t_2=T_2-T_1=\sqrt[2]{\dfrac{h}{g}}(\sqrt[2]{4}-\sqrt[2]{2})$

Time in falling in height

$3h$ will be

$T_3=\sqrt[2]{\dfrac{2(3h)}{g}}$

so the time in falling third

$h$ will be

$t_3=T_3-(T_1+T_2)=\sqrt[2]{\dfrac{h}{g}}(\sqrt[2]{6}-\sqrt[2]{4}-\sqrt[2]{2})$

the ratio will be

$t_1:t_2:t_3=\sqrt[2]{2}:\sqrt[2]{2}(\sqrt[2]{2}-1):\sqrt[2]{2}(\sqrt[2]{3}-\sqrt[2]{2}-1)$

$\sqrt[2]{2}$ is common in all so can be cancelled so the answer is option D.

The intial velocity of a body moving along a straight line is 7

$m{s^{ - 1}}$ It has a uniform acceleration of 4

$m{s^{ - 1}}$ . The distance covered by the body in the 5th second of its motion is :

0%
25m
0%
12.5m
0%
20m
0%
30m

A freely falling body, fall from

$45\ m$ , covers of total distance in

$3rd$ second.

0%
$8$ %
0%
$35.5$ %
0%
$55.55$ %
0%
$45.5$ %

Explanation

Total height

$h = 45\ m$ .

apply second equation of motion-

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

$u = 0$ ; freely falling so initial velocity is zero.

$a = 10\ ms^{-2}$

$\Rightarrow 45 = 0 + \dfrac{1}{2} \times 10 \times t^2$

$\Rightarrow t^2 = 9$

$\Rightarrow t = 3\ s$

Now, Distance covered in initial two second.

$h_2 = ut + \dfrac{1}{2}at^2$

$\Rightarrow h_2 = 0 + \dfrac{1}{2} \times 10 \times 2^2$

$\Rightarrow h_2 = 20\ m$

So, distnace in

$3rd$ second-

$S^{3rd} = 45-20$

$\Rightarrow S^{3rd} = 25\ m$

Total percentage -

$S^{3rd} = \dfrac{ 25 }{ 45 } \times 100$

$\Rightarrow S^{3rd} = 55.55 %$

A ball released from a height '

$h$ ' touches the ground in '

$t$ ' s. After

$t/2 s$ since dropping, the height of the body from the ground

0%
$\dfrac { h }{ 2 }$ from the ground
0%
$\dfrac { h }{ 4 }$ from the ground
0%
$\dfrac { 3h }{ 4 }$ from the ground
0%
$\dfrac { 3h }{ 2 }$ from the ground

A body freely falling from a height h describe

$\dfrac {11h}{36}$ in last esecond of its fall. the height h is

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

0%
$125 m$
0%
$180 m$
0%
$360 m$
0%
$90 m$

A stone is thrown vertically upward with a speed of

$49\ ms^{-1}$ . Find the velocity of the stone one second before it reaches the maximum height.

0%
$4.9 ms^{-1}$
0%
$9.8 ms^{-1}$
0%
$13.6 ms^{-1}$
0%
$19.6 ms^{-1}$

Explanation

Initial velocity of the stone is

$u= 49\ m/s$

Acceleration:

$a=g=9.8\ ms^{-2}$

While ascending, the stone will deccelerate and finally its velocity will become zero at the maximum height.

Let

$t$ is the time taken by the stone to reach the maximum height.

Using the equation

$v=u-gt$

$t=\dfrac{v-u}{-g}=\dfrac{0-49}{-9.8}=5\ sec$

We need to find the velocity of the stone at

$t=5-1=4\ sec$

Again, Using the equation

$v=u-gt$

$v=49-(9.8\times4)=49-39.2=9.8\ m/s$

A ball is dropped from a balloon moving upwards with velocity

$10\ m/s$ at a height of

$20\ m$ from the ground. The distance traveled by the ball before reaching the ground is: (Take

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

0%
$40\ m$
0%
$10\ m$
0%
$30\ m$
0%
$20\ m$

Explanation

Initial velocity of the ball,

$u = 10\ m/s$
Height of the balloon

$h = 20\ m$
From A to B

Final velocity,

$v = 0\ m/s$
Let

$x$ be the distance travelled by the ball till i's velocity

$v = 0$

Using third equation of motion,

$2ax = v^{2} - u^{2}$

$-2gx = v^{2} - u^{2}$ (

$a = -g$ downward)

$-2 \times 10 \times x = 0 - 10^{2}$

$x = \dfrac{100}{20} = 5\ m$

$\therefore$ Total distance

$D$ travelled by the balloon

$D = h + x + x = h + 2x$

$D = 20 + (2 \times 5)$

$D = 30\ m$

2093321_1295037_ans_27a3b9edfca2432fa4bdcaaa42be460f.png

From the top of a tower, a stone is thrown up and it reaches the ground in time

$t_1$ . A second stone is thrown down with the same speed and it reaches the ground in time

$t_2$ . A third stone is released from rest and it reaches the ground in time

$t_3$ .

0%
$t_3 = \frac{1}{2} (t_1 +t_2)$
0%
$t_3 = \sqrt{t_1 t_2}$
0%
$\frac{1}{t^3} = \frac{1}{t_2} - \frac{1}{t_1}$
0%
$t_3^2 = t_1^2 - t_2^2$

Starting from rest, an athlete attains a speed of

$72 km/h$ in

$5$ seconds undergoing uniform acceleration. If he runs in a straight line, the distance travelled by him in this time interval is

0%
$45 m$
0%
$90 m$
0%
$50 m$
0%
$108 m$

Which of the following is a measurement of velocity ?

0%
$15$ $m$ $east$
0%
$35$ $km/h$ $north$
0%
$18$ $m/s$ $^2$
0%
$50$ $km/h$

Explanation

Velocity is defined as speed plus direction so unit of velocity should be unit of speed plus direction.So option B has unit of speed

$km/h$ and direction as

$north$ .Hence it is correct option.

Rajdhini express moves at a speed of

$150\ km/h$ . How long will it take to cover a distance of

$15\ km$ ?

0%
$0.1\ h$
0%
$1.125\ h$
0%
$2.125\ h$
0%
$3.125\ h$

Explanation

Given,

Speed

$v=150\ km/h$

distance cover = displacement

$d=15\ km$ (Assume express is not changing its direction)

$Velocity\ (V)=\dfrac{Displacement\ (d)}{time\ (t)}$

$t=\dfrac dV=\dfrac{15}{150}=0.1\ hr$

Time taken

$t=0.1\ hr$ Option A

A body moves with initial velocity

$10ms^{-1}$ . If it covers a distance of

$20m$ in

$2s$ , then acceleration of the body is

0%
Zero
0%
$10ms^{-2}$
0%
$5ms^{-2}$
0%
$2ms^{-2}$

Explanation

Suppose the acceleration is

$a$ then using the second equation of motion,

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

and putting

$t=2sec$ and

$u=10m/s$

we get

$20=10\times 2+a\times 2$

$a=0$ that is

$uniform$ motion.

For a particle in uniform circular motion:

0%
both velocity and acceleration are constant
0%
acceleration and speed are constant but velocity changes
0%
both acceleration and velocity changes
0%
Speed is constant

A ball is dropped from the top of a building

$100 \ m$ high. At the same instant another ball is thrown upwards with a velocity of

$40\ ms^{-1}$ from the bottom of the building. The two balls will meet after.

0%
$5 s$
0%
$2.5 s$
0%
$2 s$
0%
$3 s$

Explanation

Total height

$=100\,m=S_1+S_2$

Using equation of motion,

For first ball

$S_1=ut+\dfrac 12gt^2$

where,

$S=$ Distance

$u=$ Initial velocity

$g=$ acceleration due to gravity

$t=$ time

$S_1=0-\dfrac 12 gt^2$

For second ball

$S_2=40t+\dfrac 12 gt^2$

$S_2=40t-S_1$ .......(i)

Putting the value of

$S_1$ in equation (i)

$S_2+S_1=40t$

$100=40t$

$t=2.5\,sec$

Hence, two balls will meet after

$2.5\,sec$

A body performing uniform circular motion in a horizontal plane has a constant

0%
Velocity
0%
Acceleration
0%
Kinetic energy
0%
Displacement

Can a body have a constant speed but variable velocity?

0%
Yes
0%
No
0%
May be
0%
None of these

$350\,m$ long train passes over a

$250\,m$ long bridge at a speed of

$54\,km/h$ . How long will it takes to cross the bridge?

0%
$40\,minutes$
0%
$2/3\,minutes$
0%
$2/3\,hr$
0%
$1/90 s$

Explanation

To cross bridge completely train has to cover distance, which is equal to sum of the length of train and bridge.

Distance to cover

$, S = 250 +350 = 600 m$

Speed of train

$, v = 54km/hr = 54 \times \frac{1000m}{3600s}=15m/s$

Therefore,

Time taken by trian to cross bridge

$, t = \frac{Distance}{Speed}$

$t = \frac{600}{15} =40sec$

$Or \ t= \frac{40}{60} min= \frac{2}{3}min$

Hence option B is correct

The figure represents the velocity-time graph of body moving in a straight line. How much distance does it travel during the last 10 seconds?

0%
20 m
0%
100 m
0%
50 m
0%
200 m

Explanation

Distance travelled in last

$10$ sec, is

$d = velocity \times time$ , which is nothing but area under the curve between the limit

$t = 10s$ to

$t = 20s$

So, area of triangle is

$= \dfrac {base \times height}{2} = \dfrac {(20 - 10)\times 20}{2} = 100\ m$

Hence, correct option is (B).

The displacement-time graph for two particles A and B is as follows. The ratio

$\dfrac{V_A}{V_B}$ is

0%
$1 : 2$
0%
$1 : \sqrt{3}$
0%
$\sqrt{3} : 1$
0%
$1 : 3$

Which of the following remains constant in UCM?

0%
Speed
0%
Kinetic energy
0%
Angular speed
0%
All of these

A body is released from the top of tower

$h$ metre. It takes time

$T$ seconds to reach the ground.Then the height of the body at time

$\frac { T } { 2 }$

0%
At $d\frac { h } { 2 }$ m from the ground
0%
At $\dfrac { h } { 4 }$ m from the ground
0%
At $\dfrac { 3h } { 4 }$ m from the ground
0%
Depends upon mass and volume of the body

When a body is accelerated: (i) its velocity always changes (ii) its speed always changes (iii) its direction always changes (iv) its speed may or may not change Which of the following is correct?

0%
(i) and (ii)
0%
(i) and (iv)
0%
(i) only
0%
(ii) and (iii)

A body is moving with uniform velocity of

$8\ ms^{-1}$ . When the body just crossed another body, the second one starts and moves with uniform acceleration of

$4\ ms^{-2}$ . The time after which two bodies meet will be:

0%
$2\ s$
0%
$4\ s$
0%
$6\ s$
0%
$8\ s$

A particle is moving with velocity of

$4\ ms^{-1}$ along positive

$x-$ direction. An acceleration of

$1\ ms^{-2}$ is aceted on the particle along negative

$x-$ direction. Find the distance traveled by the particle in

$10\ s$ .

0%
$10\ m$
0%
$26\ m$
0%
$16\ m$
0%
$8\ m$

Explanation

In this case, velocity and acceleration are acting in opposite direction and velocity will be zero after some time and the particle will change its direction of motion.

Apply first equation of motion-

$v = u + at$

When velocity is getting zero

$0 = 4 + (-1)t$

$t = 4\ sec$

Distance travelled in

$10\ s =$ (Distance travelled in

$4s$ )+ (Distance travelled in

$6s$ )

Apply second equation of motion-

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

$\Rightarrow s = \left[4\times 4-\dfrac {1}{2}\times 1\times (4)^2 \right]+\left[0\times 6-\dfrac {1}{2}\times 1\times (6)^2 \right]$

$=26\ m$

The velocity-time graph of a body is shown in Fig.4.The displacement of the body in 8 s is

0%
9 $m$
0%
12 $m$
0%
10 $m$
0%
28 $m$

Explanation

Displacement = area under graph

$Area=\quad 2\times 2+\dfrac { 1 }{ 2 } \left( 2+6 \right) \times 1+\dfrac { 1 }{ 2 } \times 1\times 6-\dfrac { 1 }{ 2 } +1\times 6-1\times 6+2\times 4$

$=4+4+3-3-6+8=10m$

$Area=10m=displacement \ in \ 8sec.$

Which of the following is not an example of acceleration?

0%
A person jogging at $3\ m/s$ along a winding path
0%
A car stopping at a stop sign
0%
A cheetah running $27\ m/s$ east
0%
A plane taking off

CBSE Questions for Class 9 Physics Motion Quiz 11 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

CBSE Questions for Class 9 Physics Motion Quiz 11 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

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