Using the table given below where the values of velocity at the end of t seconds for a body under linear motion are given $$V (m\:s^{-1})$$ 0 6 12 24 30 36 42 $$t (s)$$ 0 2 4 8 10 12 14 What can be concluded about the motion of the body?

It moves with uniform acceleration

MCQ Questions for Class 9 Physics Motion Quiz 12

The following figure gives the movement of an object. Select the correct statement from the given choices.

0%
The total distance travelled by the object is $975\ m$
0%
The maximum acceleration of the object is $2\ ms^{-2}$
0%
The maximum deceleration happened between $25th$ and $85th$ seconds
0%
The object was at rest between $10th$ and $15th$ seconds
0%
At $40th$ second, the speed of object was decelerating

Speed is to velocity as:

0%
Centimetre is to metre
0%
Force is to torque
0%
Velocity is to acceleration
0%
Distance is to displacement

The height of the body after 5 s from the ground is

$(g=9.8m/s^2).$

0%
8 m
0%
12 m
0%
18 m
0%
24 m

Explanation

Initial velocity will be same as balloon velocity from which it is dropped,

Considering upward direction as positive

$u=4m/s$ Body was going up initially with balloon with speed of

$4m/s$

$a=-9.8m/s^2$ Acceleration due to gravity downwards will be taken negative.

$t=5s$ Initial height =

$120.5m$

Newton Second Law of motion

$s=ut+\dfrac{1}{2}at^2=4\times5-\dfrac{1}{2}\times9.8\times5^2$

$=20-122.5=-102.5m$
This shows that the body is

$102.5 m$ below the initial position, i.e., height of the body

$= 120.5 - 102.5 = 18 m$

The speed-time graph of a body is straight line parallel to time axis. The body has:

0%
uniform acceleration
0%
uniform speed
0%
variable speed
0%
variable acceleration

Explanation

If the speed-time graph is parallel to the time axis, then this means that the body has the same speed at all the time.

Thus, speed

$=$ constant

Hence the body has uniform speed and thus option B is correct.

1628542_1773478_ans_8fc22fdb5d9643f9acadc512e9d5add2.png

Two bodies , one held

$1 \, m$ vertically above the other , are released simultaneously and fall freely under gravity . After

$2$ second , the relative separation of the bodies will be :

0%
$4.9\, m$
0%
$9.8\,m$
0%
$19.6\,m$
0%
$1 \,m$

Explanation

Now, initial separation

$= 1\,m$

Body $1$ :

Initial Speed $= 0$

Acceleration $= 9.8 \,ms^{-2}$

Time $= 2$ seconds

Now, we know that,

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

So, $s_1 = \dfrac{1}{2} \times 9.8 \times 4$

$s_1 = 19.6 \, m$

Body $2$ :

Initial Speed $= 0$

Acceleration $= 9.8\, ms^{-2}$

Time $= 2$ seconds

Now, we know that,

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

So,

$s_2 = \dfrac{1}{2} \times 9.8 \times 4$

$s_2 = 19.6 \, m$

Since

$s_1 = s_2$ so , final separation is also

$1\,m$ .

A bus travels 54 km in 90 minutes. The speed of the bus is?

0%
0.6m/s
0%
10m/s
0%
5.4m/s
0%
3.6 m/s

If we denote speed by

$S$ , distance by

$D$ and time by

$T$ the relationship these equation is?

0%
$S= D \times T$
0%
$T= \frac {S}{D}$
0%
$S = \frac {1}{T} \times D$
0%
$S= \frac {T}{D}$

Explanation

$speed = \dfrac{distance}{Time}$

Four cars

$A, B, C$ and

$D$ are moving on a levelled road. Their distance versus time graphs are shown in Fig

$8.2$ . Choose the correct statement

0%
Car $A$ is faster than car $D$ .
0%
Car $B$ is the slowest.
0%
Car $D$ is faster than car $C$ .
0%
Car $C$ is the slowest.

Explanation

Instantaneous Speed of car

$v=\dfrac{dx}{dt}$

$dx=v\ dt$

$x=v\ t$ ............(1)

For faster car velocity will be more and it will cover more distance relative to others. at the same time slower car will travel less distance.

Form the above graph car be is slowest.

Option B

A body is thrown vertically upward with velocity

$u$ , the greatest height

$h$ to which it will rise is,

0%
$\dfrac{u}{g}$
0%
$\dfrac{u^{2}}{2g}$
0%
$\dfrac{u^{2}}{g}$
0%
$\dfrac{u}{2g}$

Explanation

Given, initial velocity = u, height = h and a = g (acceleration due to gravity)

At the highest point, final velocity becomes zero i.e., v = 0

From, third equation of motion,

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

$0=u^2-2gh$

$u^2=2gh$

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

The correct symbol to represent the speed of an object is

0%
$5m/s$
0%
$5mp$
0%
$5 m/ sec^{-1}$
0%
$5 s/m$

Explanation

Velocity is rate of change of displacement with respect to time.

Velocity

$V=\dfrac DT$

So, unit of velocity

$=\dfrac ms$

Hence Option : A

A black paper absorbs light of all the colours and reflects none.

0%
True
0%
False

Explanation

From conversion from

$ms^{-1}$ to

$kmh^{-1}$ we get,

$1km=1000m$ and

$1hr=3600sec$

$ms^{-1}$ =

$\dfrac{5m}{sec}$ =

$\dfrac{5 \times 10^{-3}km}{1/3600hr}$ =

$0.005 \times 3600$ =

$18kmh^{-1}$

Hence,the speed 5

$ms^{-1}$ is less than 25 km

$h^{-1}$ .

State whether the following statement is true or false.
In a uniform circular motion, the speed continuously changes because of the direction of motion changes.

0%
True
0%
False

$18\ km\ h^{-1}$ is equal to :

0%
$10\ ms^{-1}$
0%
$5\ ms^{-1}$
0%
$18\ ms^{-1}$
0%
$1.8\ ms^{-1}$

Explanation

$5\ ms^{-1}$

As,

$1km=1000m$

and

$1\ hour = 60\times 60\ seconds$
Converting

$\dfrac{18 \times 1000}{60\times 60}=5\ ms^{-1}$

A body falls from rest, its velocity at the end of first second is

$( g = 32 \,ft / s^2)$

0%
$16 \, ft / s$
0%
$32 \, ft / s$
0%
$64 \, ft / s$
0%
$24 \, ft / s$

Explanation

Since the body falls from rest, the initial velocity,

$u=0 ft/s.$

$g= 32\ ft/s^2$

The velocity attained by body in time

$t = 1\ s$ is given by :

$\nu = u +g t = 0 +32 \times 1 = 32 \, ft/s$

Option (B) is correct.

A body thrown with an initial speed of

$96\, ft / sec$ reaches the ground after

$( g = 32 ft / sec^2 )$

0%
$3 \,sec$
0%
$6 \,sec$
0%
$12 \,sec$
0%
$8 \,sec$

Explanation

Time of flight

$T = \dfrac{2u}{g}$

$T\ = \dfrac{2 \times 96}{32} = 6 \, sec$

The variation of velocity of a particle with time moving along a straight line is shown in the figure. The distance traveled by the particle in

$4\ s$ is :

0%
$60\,m$
0%
$55 \,m$
0%
$25 \,m$
0%
$30\,m$

Explanation

$Distance$ =

$Area\ under\ v-t\ graph$

$= A_1 + A_2 + A_3 + A_4$

$= \dfrac{1}{2} \times 1 \times 20 + (20 \times 1) + \dfrac{1}{2} (20 + 10) \times 1 + (10 \times 1)$

$= 10 + 20 + 15 + 10 = 55 m$
1705383_1812909_ans_93e9b9ad1734434ebddc472bbf6dd455.png

Time taken by an object falling from rest to cover the height of

$h_1$ and

$h_2$ is respectively

$t_1$ and

$t_2$ then the ratio of

$t_1$ to

$t_2$ is

0%
$h_1 : h_2$
0%
$\sqrt{h_1} : \sqrt{h_2}$
0%
$h_1 : 2 h_2$
0%
$2h : h$

Explanation

By 2nd equation of motion,

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

Since

$u=0 \ and \ a = g$ we have:

For $s=h_1, t = t_1$

$h_1=\dfrac{1}{2}gt_1^2 \implies t_1=\sqrt{\dfrac{2h_1}{g}} . . . . . . (i)$

For $s=h_2, t = t_2$

$h_2=\dfrac{1}{2}gt_2^2 \implies t_2=\sqrt{\dfrac{2h_2}{g}} . . . . . . (ii)$

By dividing (i) by (ii)

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

An object moving in upward direction opposite to the gravitational force of earth performs

0%
accelerated motion
0%
motion with constant velocity
0%
oscillations
0%
retarded motion

Explanation

Answer: Retarded motion.
The motion of an object vertically upwards is retarded motion. Here, the retardation is

$9.8 \ m/s^2.$ The retardation is nothing but an acceleration which acts opposite direction of velocity. So it has a tendency to slow down the object.

The horizontal straight line obtained from the distance-time graph is related to which of the following velocity ?

0%
Zero velocity
0%
Constant velocity
0%
Increasing velocity
0%
Decreasing velocity

Four alternatives are given to each of the following incomplete statements/questions, choose the right answer.
Uniform circular motion is called continuously accelerated motion mainly because

0%
Direction of motion changes
0%
Speed remains the same
0%
Velocity remains the same
0%
Direction of motion does not change

Figure represents the displacement-time graph of motion of two cars A and B. Find the velocities of car A and car B.

0%
$v_A = 20 km h^{-1}, v_B = 10 km h^{-1}$
0%
$v_A = 10 km h^{-1}, v_B = 20 km h^{-1}$
0%
$v_A = 20 km h^{-1}, v_B = 40 km h^{-1}$
0%
$v_A = 40 km h^{-1}, v_B = 20 km h^{-1}$

Explanation

For the time interval of

$4 h$ ,

Displacement of car A,

$D_A = 160 km$
Velocity of car A,

$v_A = \dfrac{160km}{4 h} = 40 km/h$

Displacement of car B,

$D_B = 120 km-40 km = 80 km$
Velocity of car B,

$v_B = \dfrac{80 km}{4 h} = 20 km/h$

In the figure, the displacement time graph of a body is shown at different time-intervals. Calculate the velocity of the body as it moves for

$(i)\ 0 \ to\ 5 s$ ,

$(ii)\ 5 s \ to\ 7 s$ and

$(iii)\ 7 s\ to\ 9 s$ .

0%
$(i) \ 1.2 m s^{-1} (ii)\ 0 m s^{-1} (iii) \ 1 m s^{-1}$
0%
$(i) \ 0.6 m s^{-1} (ii) \ 0 m s^{-1} (iii)\ 2 m s^{-1}$
0%
$(i) \ 0 m s^{-1} (ii) \ 0.6 m s^{-1} (iii) \ 4 m s^{-1}$
0%
$(i)\ 0.3 m s^{-1} (ii) \ 1 m s^{-1} (iii)\ 1.2 m s^{-1}$

Explanation

$Velocity, v = \dfrac{Displacement, d}{Time, t}$

$(i)$

$v = \dfrac{3m}{5s} = 0.6 m/s$

$(ii)$

$v = \dfrac{0m}{2s} = 0 m/s$

$(iii)$

$v = \dfrac{7m-3m}{2s} = 2 m/s$

Two masses of mass

$40\ gm$ and

$30\ gm$ are connected across a pulley with the help of light string, then the acceleration of the system is

0%
$4.9 m/s^{2}$
0%
$2.94 m/s^{2}$
0%
$1.4 m/s^{2}$
0%
$9.8 m/s^{2}$

Explanation

The resultant forces for the given masses are:

$0.04g-T = 0.04a$ ...(1)
and

$T -0.03g = 0.03a$ ...(2)
Adding (1) and (2), we have

$\therefore 0.04g-(0.03a + 0.03g) = 0.04a$

$0.01g=0.07a$ or $a=\frac{0.098}{.07}$ or $a = 1.4$

The acceleration of the system $a = 1.4 m/s^2$

A lift performs the first part of its ascent with uniform acceleration,

$a$ , and the remaining with uniform retardation,

$2a$ . If,

$t$ , is the time of ascent, the depth of the shaft is

0%
$\displaystyle \dfrac{{at}^{2}}{4}$
0%
$\displaystyle \dfrac{{at}^{2}}{3}$
0%
$\displaystyle \dfrac{{at}^{2}}{2}$
0%
$\displaystyle \dfrac{{at}^{2}}{8}$

Explanation

We have

$t_1=\dfrac{v}{a}$ and

$t_2=\dfrac{v}{2a}$
Adding both times, we get

$t=\dfrac{3v}{2a}$ ...... (1)
or

$h=h_1+h_2$

$h=\dfrac{v^2}{2a}+\dfrac{v^2}{4a}$

$h=\dfrac{3v^2}{4a}$ ...... (2)

From (1) and (2), we get

$h=\dfrac{at^2}{3}$

Using the table given below where the values of velocity at the end of t seconds for a body under linear motion are given

$V (m\:s^{-1})$	0		6	12	24	30	36	42
$t (s)$	0		2	4	8	10	12	14

What can be concluded about the motion of the body?

0%
It moves with uniform speed.
0%
It moves with uniform motion
0%
It moves with uniform velocity
0%
It moves with uniform acceleration

What is the minimum height above the ground at which the rocketeer should catch the student?

0%
$92.1\ m$
0%
$460.9\ m$
0%
$78.8\ m$
0%
$82.3\ m$

Explanation

$Let\quad height\quad at\quad which\quad student\quad is\quad caught\quad is\quad h\\ velocity\quad at\quad this\quad point\quad is\quad \sqrt { 2g(553-h) } \\ when\quad the\quad student\quad comes\quad to\quad ground\quad his\quad velocity\quad is\quad zero\\ { v }^{ 2 }-{ u }^{ 2 }=2as\\ 0-2g(553-h)=2(-5g)h\\ \Rightarrow 553-h=5h\Rightarrow h=92.1m$

A smooth track of incline of length

$l$ is joined smoothly with circular track of radius

$R$ . A mass of

$m$ kg is projected up from the bottom of the inclined plane. The minimum speed of the mass to reach the top of the track is given by,

$v=$

0%
${ \left[ 2g(l\cos { \theta } +R)(1+\cos { \theta } ) \right] }^{ 1/2 }$
0%
${ \left( 2gl\sin { \theta } +R \right) }^{ 1/2 }$
0%
${ \left[ 2g\{ l\sin { \theta } +R(1-\cos { \theta } )\} \right] }^{ 1/2 }$
0%
${ \left( 2gl\cos { \theta } +R \right) }^{ 1/2 }$

Explanation

When the mass reach to the top , the final velocity

$=0$ and height,

$H=h+h'$
or,

$H=l\sin\theta+R(1-\cos\theta)$
using,

$v^2-u^2=2aS$ , we get

$0-u^2=2(-g)H \Rightarrow u^2=2g[l\sin\theta+R(1-\cos\theta)]$
or

$u=\left[2g\{l\sin\theta+R(1-\cos\theta)\}\right]^{1/2}$

A particle experiences a net force that is always parallel to y-axis. If it moves along the curve

$xy=a^2$ where

$a$ is a constant and the magnitude of the force is

$y^n$ , then

$n$ is

0%
1
0%
2
0%
3
0%
4

Explanation

$\vec{F} = y^{n} \hat{j}$

$\vec{S} = \dfrac{a^2}{y} \hat{i} + y \hat{j}$

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

$=> v^2 - u^2 = \dfrac {2FS} {m}$

$=> v^2 - u^2 = \dfrac {2y^{n+1}} {m}$

The difference of the two square terms is expressed as a constant multiplied by (n+1)th power of third term.

So (n+1) = 2

=> n = 1

With what speed in miles/hour (

$1 m/s = 2.23 mi/h$ ) must an object be thrown to reach a height of

$91.5 m$ (equivalent to one football field)? Assume negligible air resistance.

0%
$94.4 mi/h$
0%
$84.4 mi/h$
0%
$74.4 mi/h$
0%
$94 mi/h$

Explanation

As we assume negligible air resistance, so the acceleration of the object is due to gravity, i.e.

$a=-g=-9.8 m/s^2$

where minus sign for motion against gravity.

When the object reaches top most height, the final velocity will be zero i.e

$v=0 \ m/s$

Maximum height reached

$S = 91.5 \ m$

Let

$u$ be the initial velocity of the object.

Using formula

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

$\therefore$

$0^2-u^2=2(-9.8)(91.5)$

$\implies$

$u=42.35 m/s=42.35\times 2.23 mi/h=94.4 mi /h$

Which of the following statements is correct for a particle travelling with a constant speed?

0%
Its position remains constant as time passes.
0%
It covers equal distances in unequal time intervals.
0%
Its acceleration is zero.
0%
It does not change its direction of motion.

A race car accelerates uniformly from

$18.5 m/s$ to

$46.1 m/s$ in

$2.47$ seconds. Determine the distance traveled by the car. (in m)

0%
$79.8 m$
0%
$79 m$
0%
$7.8 m$
0%
$9.8 m$

Explanation

Here, initial velocity is

$u=18.5 m/s$ and final velocity is

$v=46.1 m/s$

Time taken

$t=2.47 s$

$a$ be the acceleration of the car.

Using

$v=u+at$ ,

$46.1=18.5+a(2.47)$

$\implies$

$a=11.17 m/s^2$

$S$ be the distance traveled by car.

Using formula

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

$(46.1)^2-(18.5)^2=2(11.17)S$

$\implies$

$S=79.8m$

An engineer is designing the runway for an airport of the planes that will use the airport, the lowest acceleration rate is likely to be

$3 m/s^2$ . The takeoff speed for this plane will be

$65 m/s$ . Assuming this minimum acceleration, what is the minimum allowed length for the runway? (in m)

0%
$704$
0%
$70$
0%
$705$
0%
$707$

Explanation

Given, initial velocity,

$u=0 m/s$ and final velocity

$v=65 m/s$

Acceleration

$a=3 m/s^2$

$d$ be the allowed length for the runway.

Using formula

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

$(65)^2-0^2=2(3)d$

$\implies$

$d=65^2/6=704.16 \sim 704 m$

The driver of a train travelling at

$115\ km\ h^{-1}$ seen on the same track,

$100\ m$ in front of him, a slow train travelling in the same direction at

$25\ km\ h^{-1}$ . The least retardation that must be applied to faster train to avoid a collision is

0%
$25\ ms^{-2}$
0%
$50\ ms^{-2}$
0%
$75\ ms^{-2}$
0%
$3.125\ ms^{-2}$

Explanation

Relative velocity of faster train w.r.t. slower train

$u =115 - 25= 90\ km\ h^{-1}$

$u= 90\times \dfrac {5}{18} ms^{-1} = 25\ ms^{-1}$
Using

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

$0^{2} - 25^{2} = 2\times a\times 100$
or

$200a = -625 \Rightarrow a = -\dfrac {625}{200} ms^{-2}$
or

$a = -3.125\ ms^{-2}$ .

The graph below shows the distance travelled and the time taken by four cars.
Which car travelled the slowest?

0%
Car $1$
0%
Car $2$
0%
Car $3$
0%
Car $4$

Explanation

The slope of the distance-time graph represents the velocity.

The slope for the car

$4$ is minimum, so its speed is also minimum.

A stone is thrown vertically upwards. When the stone is at a height equal to half of its maximum height its speed will be 10 m/s, then the maximum height attained by the stone is (Take g =10

$m/s^2$ )

0%
3 m
0%
15 m
0%
1 m
0%
10 m

Explanation

Let

$u$ be the initial velocity and

$h$ be the maximum height attained by the stone.

$v^2_1 = u^2 - 2gh$ ,

$(10)^2 = u^2 - 2\times 10 \times \frac {h}{2}$

$100=u^2 - 10h$ ....(i)
Again at height

$h$ ,

$v^2_2 = u^2 - 2gh$

$(0)^2=u^2-2 \times 10\times h$

$u^2 = 20\, h$ ... (ii)
So, from Eqs. (i) and (ii) we have

$100 = 10h$

$h=10\, m$

The figure shows the displacement time graph of a particle moving on a straight line path. What is the magnitude of average velocity of the particle over 10 s?

0%
$2\;ms^{-1}$
0%
$4\;ms^{-1}$
0%
$6\;ms^{-1}$
0%
$8\;ms^{-1}$

Explanation

Net displacement of the particle in 10 s,

$S = BC - OA = 40 -60 = -20$ m

Total time taken

$t = 10$ s

$\therefore$ Average velocity of the particle

$v_{avg} = \dfrac{S}{t} = \dfrac{-20}{10} = -2 m/s$

Thus magnitude of average velocity of the particle is

$2 m/s$ .

A car, moving at 1.5

$m{s}^{-1}$ applies brakes and comes to rest in

$2 s$ . If the same car travels at double the speed, what time would it take to come to rest after applying brakes?

0%
$4s$
0%
$8s$
0%
$2s$
0%
$3s$

Explanation

Given,

Initial velocity

$u_1=1.5ms^{-1}$
Final velocity

$v_1=0$
Time taken

$t_1=2s$

According to first equation of motion,

$v_1=u_1+a_1t_1$

$0=1.5+2a_1$

$a_1=\dfrac{-1.5}{2} = -0.75\ ms^{-2}$

Now, for the second case,

Initial velocity

$u_2=3ms^{-1}$

Final velocity

$v_2=0$

The braking system is still the same. So it provides the same retarding acceleration of

$-0.75 \ ms^{-1}$

So acceleration

$a_2$ is same as

$a_1=-0.75 \ ms^{-1}$

Using first equation of motion again,

$v_2=u_2+a_1t_2$

$0=3-0.75\times t_2$

$t=\dfrac{3}{0.75}=4s$

The particle is moving with constant speed

0%
In graphs (i) and (iii)
0%
In graphs (i) and (iv)
0%
In graphs (i) and (ii)
0%
In graphs (i)

Two cyclists, k km apart, and starting at the same time, would be together In r hours if they travelled in the same direction, but would pass each other In t hours if they travelled in opposite directions. The ratio of the speed of the faster cyclist to that of the slower is ____

0%
$\frac{r+t}{r-t}$
0%
$\frac{r}{r-t}$
0%
$\frac{r+t}{r}$
0%
$\frac{r}{t}$

The acceleration will be positive in:

0%
(I) and (III)
0%
(I) and (IV)
0%
(II) and (IV)
0%
None of these

A spy plane is being tracked by a radar. At

$t=0$ , its position is reported as

$(100 m,\, 200 m, \,1000 m)$ .

$130 s$ later, its position is reported to be

$(2500 m,\, 1200 m,\, 1000 m)$ . Find a unit vector in the direction of plane velocity and the magnitude of its average velocity.

0%
$20ms^{-1}$ ; $\dfrac{22\hat{i}+5\hat{j}}{13}$
0%
$20ms^{-1}$ ; $\dfrac{12\hat{i}+5\hat{j}}{13}$
0%
$30ms^{-1}$ ; $\dfrac{12\hat{i}+5\hat{j}}{13}$
0%
$20ms^{-1}$ ; $\dfrac{12\hat{i}+6\hat{j}}{13}$

Explanation

Velocity vector is in direction of displacement vector.

$\vec { d } =\left( 2500-100 \right) \hat i +\left( 1200-200 \right) \hat { j } +\left( 1000-1000 \right) \hat { k }$

$=2400\hat i +1000\hat { j }$

$\vec { V } =\dfrac { \vec { d } }{ t } =\dfrac { 2400 }{ 130 } \hat i +\dfrac { 1000 }{ 130 } \hat { j } =\dfrac { 240 }{ 13 } \hat i +\dfrac { 100 }{ 13 } \hat { j }$

$\hat { V } =\dfrac { \vec { V } }{ \left| V \right| } =\dfrac { \dfrac { 240 }{ 13 } \hat i +\dfrac { 100 }{ 13 } \hat { j } }{ \sqrt { { \left( \dfrac { 240 }{ 13 } \right) }^{ 2 }+{ \left( \dfrac { 100 }{ 13 } \right) }^{ 2 } } } =\dfrac { 240\hat i +100\hat { j } }{ \sqrt { { \left( 240 \right) }^{ 2 }+{ \left( 100 \right) }^{ 2 } } }$

$=\dfrac { 240 }{ 260 } \hat i +\dfrac { 100 }{ 260 } \hat { j } =\dfrac { 12 }{ 13 } \hat i +\dfrac { 5 }{ 13 } \hat { j }$

$\left| \vec { V } \right| ={ \left( { \left( \dfrac { 240 }{ 13 } \right) }^{ 2 }+{ \left( \dfrac { 100 }{ 13 } \right) }^{ 2 } \right) }^{ 1/2 }=\sqrt { 400 } =20{ ms }^{ -1 }$

Speed - distance graph of an object which comes to rest suddenly is

Consider the graph given.
The distance travelled by an object in a given direction as time increases is given by the given graph.
In this context which of the following statements is correct?

0%
The velocity decreases during DE
0%
The velocities during DE and BC are in opposite directions
0%
Total distance travelled is AF
0%
During CD the body is moving with uniform velocity

A body was initially moving with

$10\ m/sec$ and it starts acceleration with

$2\ m/sec^{2}$ , the distance covered by it in

$10\ sec$ will be

0%
$200\ m$
0%
$100\ m$
0%
$150\ m$
0%
$250\ m$

Explanation

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

$s=10\times 10+\dfrac{1}{2}\times 2\times 10^2=200m$

A train is travelling at a speed of

$108\ km/h$ . Brakes are applied so as to produce a uniform retardation of

$1\ m/s^{2}$ . Find how far the train will go before it is brought to rest :

0%
$800\ m$
0%
$450\ m$
0%
$1000\ m$
0%
$1050\ m$

Explanation

Given,

$u=108km/h=\dfrac{108\times 1000}{60\times 60}=30m/s$

$a=-1m/s^2$

$v=0m/s$

From 3rd equation of motion,

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

$-2\times 1\times s=0^2-30\times 30$

$s=\dfrac{-30\times 30}{-2\times 1}$

$s=450m$

The correct option is B.

A jeep starts from rest and attains a speed of

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

$10$ minutes. The uniform acceleration will be :

0%
$4\ km\ h^{-2}$
0%
$4\ km\ s^{-2}$
0%
$66.7\ m\ s^{-2}$
0%
$1.85\ cm\ s^{-2}$

Explanation

$v=40km/h= \frac{40\times5}{18 } m/s= 11.11m/s=1111cm/s$

$t=10mins=10\times60 =600 secs$

$u=0$

$a$ =

$\dfrac{v-u}{t}=\dfrac{1111}{600}=1.85cm/s^2$

A particle has a velocity u towards east at t =Its acceleration is towards west and is constant, Let

$X_A$ and

$X_B$ be the magnitude of displacements in the first 10 seconds and the next 10 seconds.

0%
$X_A < X_B$
0%
$X_A = X_B$
0%
$X_A > X_B$
0%
the information is insufficient to decide the relation of $X_A$ and $X_B$ .

For the graph given below the ratio of average speed to the average velocity in

$( ms^{-1})$ of the particle is.

0%
$\dfrac{5}{6}$
0%
$\dfrac{5}{8}$
0%
3
0%
4

Water drops fall at regular intervals from a roof at of

$125\ m$ from the ground. At an instant when

$11th$ drop is about leave the roof and

$1st$ drop is at a height of

$45\ m$ from the ground at that instant the dist of

$10th$ drop from the roof is

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

0%
$0.40\ m$
0%
$0.20\ m$
0%
$0.80\ m$
0%
$1.2\ m$

The splash is heard

$2.05 \ s$ after the stone is dropped into a well of depth

$19.6 \ m$ . The velocity of sound is?

0%
$342$ $ms^{-1}$
0%
$372$ $ms^{-1}$
0%
$392$ $ms^{-1}$
0%
$352$ $ms^{-1}$

Explanation

If we use

$g=9.8m/{ s }^{ 2 }$ for the acceleration of gravity, and assume that there are no frictional forces to slow the stone down as it falls, then using the second equation of motion

$h=\frac { 1 }{ 2 } { gt }^{ 2 }$ gives

$t=2 \ s$ t=2s as the time at which the stone hits the water (we’re also ignoring precisely what point the stone is at

$19.6 \ m$ 19.6m above the water).

Assuming that the sound waves are emitted precisely at the moment of contact between the stone and the water, and that there is no physiological time lag in “hearing” the sound (e.g., time for auditory nerve conduction), then this means that the sound waves travel

$19.6 \ m$ 19.6m in

$t = (2.05 - 2) \ s$ 0.05s, so that their average speed is

$\dfrac { 19.6 }{ 0.05 } =392 \ m /s$

CBSE Questions for Class 9 Physics Motion Quiz 12 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

CBSE Questions for Class 9 Physics Motion Quiz 12 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

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