MCQ Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 4

A stone is allowed to fall from the top of a tower and covers half the height of the tower in the last second of its journey. The time taken by the stone to reach the foot of the tower is

0%
$\left( 2-\sqrt { 2 } \right) s$
0%
$4s$
0%
$\left( 2+\sqrt { 2 } \right) s$
0%
$\left( 2\pm \sqrt { 2 } \right) s$

Explanation

Let total distance traveled be

$h$ .
Now, Velocity at midpoint,

$v=\sqrt { 2gs } =\sqrt { gh }$

Velocity at lowest point,

$V=v=\sqrt { 2gs } =\sqrt { 2gh }$

Now,

$V=v+gt$

$(t=1s)$

Now,

$\sqrt { 2gh } =\sqrt { gh } +g$

$\Rightarrow \sqrt { h } =\dfrac { \sqrt { g } }{ \sqrt { 2 } -1 }$

Also,

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

$\Rightarrow { t }^{ 2 }-\dfrac { 2h }{ g } =0$

$t=\pm \sqrt { \dfrac { 2h }{ g } } =\pm \dfrac { \sqrt { 2 } }{ \sqrt { 2 } -1 } \\ t=2\pm \sqrt { 2 }$

A particle is thrown vertically upwards from ground. It takes time

${ t }_{ 1 }$ to reach a height

$h$ . It continues to move and takes time

${ t }_{ 2 }$ to reach the ground. Its maximum height is

0%
$\cfrac { g }{ 2 } \cfrac { { t }_{ 1 }+{ t }_{ 2 } }{ 2 }$
0%
$\cfrac { g }{ 2 } \sqrt { { { t }_{ 1 } }^{ 2 }+{ { t }_{ 2 } }^{ 2 } }$
0%
$\cfrac { g }{ 8 } { \left( { t }_{ 1 }+{ t }_{ 2 } \right) }^{ 2 }$
0%
$g({ { t }_{ 1 } }^{ 2 }+{ { t }_{ 2 } }^{ 2 })$

Explanation

Let

$u$ &

$H$ be initial velocity and maximum height respectively so,

$u=\sqrt { 2gH }$
Now, for both time

${ t }_{ 1 }\ \&\ { t }_{ 2 }$ displacement is

$h$

$h=ut-\dfrac { 1 }{ 2 } g{ t }^{ 2 }\quad$

$t^2+\dfrac{2u}{g}t+\dfrac{2h}{g}=0$

which is having two roots

$t_1$ and

$t_2$

${ t }_{ 1 }+{ t }_{ 2 }=\dfrac { 2u }{ g } \quad or\quad { t }_{ 1 }+{ t }_{ 2 }=\dfrac { 2\sqrt { 2gH } }{ g } \Rightarrow H=\dfrac { g \left( { t }_{ 1 }+{ t }_{ 2 } \right)^{2} }{ 8 }$

An aeroplane drops a parachutist. After covering a distance of

$40\ m$ , he opens the parachute and retards at

$2\ m{ s }^{ -2}$ . If he reaches the ground with a speed of

$2\ m{ s }^{ -1 }$ , he remains in the air for about

0%
$16\ s$
0%
$3\ s$
0%
$13\ s$
0%
$10\ s$

Explanation

Let

${ t }_{ 1 }$ be the time before opening of parachute.
Using

$h=ut+\dfrac { 1 }{ 2 } g{ t }^{ 2 }$ , we get,

$40=0+\dfrac { 1 }{ 2 } 9.8{ t }_{ 1 }^{ 2 }\quad \Rightarrow \quad { t }_{ 1 }=3s$
Taking

${ v }_{ 1 }$ as velocity attained after falling through

$40m$
Using

${ v }^{ 2 }={ u }^{ 2 }+2gh$ , we get

${ v }_{ 1 }^{ 2 }=0+2\left( 9.8 \right) \left( 40 \right) \quad \Rightarrow { v }_{ 1 }=28\ m/s$
Again taking

${ t }_{ 2 }$ as time taken after opening of parachute
Using

$v=u+at,$ we get

$2=28-\left( 2 \right) { t }_{ 2 }\quad \Rightarrow { t }_{ 2 }=13\ s$
Total time

$={ t }_{ 1 }+{ t }_{ 2 }=3+13=16\ s$

A driver driving a truck at a constant speed of

$20\ m{ s }^{ -1 }$ suddenly saw a parked car ahead of him by

$95\ m$ . He could apply the brake after some time to produce retardation of

$2.5\ m{ s }^{ -1 }$ .An accident was just avoided, his reaction time is

0%
$0.5 \ s$
0%
$0.75\ s$
0%
$0.8\ s$
0%
$1\ s$

Explanation

${ v }_{ rel }={ u }_{ rel }+{ a }_{ rel }t\\ \Rightarrow 0=20-2.5t\\ \Rightarrow t=8s.$
Now,

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

$s=(20\times 8)- \dfrac { 1 }{ 2 }\times 2.5 \times { (8) }^{ 2 }$

$s=160-80\\ s=80\ m$

Now, extra distance is

$15\ m$

Thus, reaction time

$=\dfrac { 15 }{ 20 } =0.75\ s$

After what time will he be able to overtake the bus?

0%
$4 sec$
0%
$8 sec$
0%
$12 sec$
0%
$16 sec$

Explanation

Let after a time $t$ , the cyclist overtake the bus. Then

$96+\displaystyle\dfrac { 1 }{ 2 } \times 2\times { t }\displaystyle^{ 2 }=20\times t$ or ${ t }\displaystyle^{ 2 }-20t+96=0$

$\therefore \quad t=\displaystyle\dfrac { 20\pm \sqrt { 400-4\times 96 } }{ 2\times 1 }$

$\quad \quad =\displaystyle\dfrac { 20\pm 4 }{ 2 } =8 sec$ and $12 sec$

But since they won't meet again after they have crossed each other once, so only $8$ seconds is the answer since they are headed in different directions.

A particle accelerates from rest at a constant rate for some time and attains a velocity of

$8 { m }/{ sec }$ . Afterwards it decelerates with the constant rate and comes to rest. If the total time taken is

$4 sec$ , the distance travelled is

0%
$32 m$
0%
$16 m$
0%
$4 m$
0%
None of the above

Explanation

Using

$v=u+at$
During acceleration,

$8=a{ t }_{ 1 } or { t }_{ 1 }=\displaystyle\dfrac { 8 }{ a }$ and during deacceleration

$0=8-a\left( 4-{ t }_{ 1 } \right)$ or

${ t }_{ 1 }=\displaystyle\dfrac { 8 }{ a }$

$\therefore \quad 8=a\left( 4-\displaystyle\dfrac { 8 }{ a } \right)$

$8=4a-8$ or

$a=4$ and

${ t }_{ 1 }={ 8 }/{ 4 }=2 sec$
Now,Distance covered during acceleration

${ s }_{ 1 }=0\times 2+\displaystyle\frac { 1 }{ 2 } \times 4{ \left( 2 \right) }\displaystyle^{ 2 }$ or

${ s }_{ 1 }=8 m$
Distance covered during deacceleration

${ s }_{ 2 }=8\times 2-\displaystyle\dfrac { 1 }{ 2 } \times 4\times { \left( 2 \right) }\displaystyle^{ 2 }$ or

${ s }_{ 2 }=8 m$
Total distance

$=\quad { s }_{ 1 }+{ s }_{ 2 }=16 m$

Which one of the following represents the displacement time graph of two objects

$A$ and

$B$ moving with zero relative speed?

Explanation

If relative speed

$= 0$ then velocity of

$A =$ velocity of

$B$ .
So displacement time graphs of A and B must have same slope (other than zero).

A particle experiences constant acceleration for

$20$ seconds after starting from rest. If it travels a distance

${ s }_{ 1 }$ in the first

$10$ seconds and distance

${ s }_{ 2 }$ in the next 10 seconds, then

0%
${ s }_{ 2 }={ s }_{ 1 }$
0%
${ s }_{ 2 }=2{ s }_{ 1 }$
0%
${ s }_{ 2 }=3{ s }_{ 1 }$
0%
${ s }_{ 2 }=4{ s }_{ 1 }$

Explanation

Let $a$ be the constant acceleration of the particle. Then $s=ut+\displaystyle\dfrac { 1 }{ 2 } a{ t }\displaystyle^{ 2 }$ or ${ s }_{ 1 }=0+\displaystyle\dfrac { 1 }{ 2 } \times a\times { \left( 10 \right) }\displaystyle^{ 2 }=50a$ and ${ s }_{ 2 }=\left[ 0+\displaystyle\dfrac { 1 }{ 2 } a{ \left( 20 \right) }^{ 2 } \right] -50a=150a$

$\therefore \quad { s }_{ 2 }=3{ s }_{ 1 }$

Alternatively:Let $a$ be constant acceleration and

$s=ut+\displaystyle\dfrac { 1 }{ 2 } a{ t }\displaystyle^{ 2 }$ , then ${ s }_{ 1 }=0+\displaystyle\dfrac { 1 }{ 2 } \times a\times 100=50a$

Velocity after 10 sec. is $v=0+10a$

So, ${ s }_{ 2 }=10a\times 10+\displaystyle\dfrac { 1 }{ 2 } a\times 100=150a\Rightarrow { s }_{ 2 }=3{ s }_{ 1 }$

Let $a$ be constant acceleration, using $s=ut+\displaystyle\dfrac { 1 }{ 2 } a{ t }\displaystyle^{ 2 }$

so distance coreved in first 10 seconds ${ s }_{ 1 }=0+\displaystyle\dfrac { 1 }{ 2 } \times a\times 100=50a$

Velocity after 10 sec. is $v=0+10a$

So, distance covered in next 10 seconds ${ s }_{ 2 }=10a\times 10+\displaystyle\dfrac { 1 }{ 2 } a\times 100=150a\Rightarrow { s }_{ 2 }=3{ s }_{ 1 }$

Free

$^{238}{U}$ nuclei kept in a train emit alpha particles. When the train is stationary and a uranium nucleus decays, a passenger measures that the separation between the alpha particles and the recoiling nucleus becomes

$x$ , in

$t$ time after the decay. If a decay takes places, when the train is moving at a uniform speed

$v$ , the distance between the alpha particle and the recoiling nucleus at a time

$t$ after the decay, as measured by the passenger will be

0%
$x+vt$
0%
$x-vt$
0%
$x$
0%
depends on the directions of the train

Explanation

Train is moving with constant velocity v.
There won't be any relative velocity change between alpha particle and uranium.
Hence, the distance will still be

$x$ .

A soldier jumps out from an aeroplane with a parachute. After dropping through a distance of

$19.6\ m$ , he opens the parachute and decelerates at the rate of

$1\ m{ s }^{ -2 }$ . If he reaches the ground with a speed of

$4.6\ m{ s }^{ -1 }$ , how long was he in air?

0%
$10\ s$
0%
$12\ s$
0%
$15\ s$
0%
$17\ s$

Explanation

Let

${ t }_{ 1 }$ be the time before opening of parachute.
Using

$h=ut+\dfrac { 1 }{ 2 } g{ t }^{ 2 }$ , we get,

$19.6=0+\dfrac { 1 }{ 2 } 9.8{ t }_{ 1 }^{ 2 }\quad \Rightarrow \quad t_1=2s$
Taking

$v_1$ as velocity attained after falling through

$19.6\ m$
Using

${ v }^{ 2 }={ u }^{ 2 }+2gh$ , we get

${ v }_{ 1 }^{ 2 }=0+2\left( 9.8 \right) \left( 19.6 \right) \quad \Rightarrow { v }_{ 1 }=19.6\ m/s$
Again taking

${ t }_{ 2 }$ as time taken after opening of parachute
Using

$v=u+at,$ we get

$4.6=19.6-\left( 1 \right) { t }_{ 2 }\quad \Rightarrow { t }_{ 2 }=15\ s$
Total time

$={ t }_{ 1 }+{ t }_{ 2 }=2+15=17\ s$

A body of mass

$3kg$ falls from the multi-storeyed building

$100m$ high and buries itself

$2m$ deep in the sand. The time of penetration is

0%
$9sec$
0%
$0.9sec$
0%
$0.09sec$
0%
$10sec$

Explanation

$V=\sqrt { 2gH } =\sqrt { 2000 } m/s\\ { v }^{ 2 }-{ u }^{ 2 }=2as\\ \Rightarrow 0-2000=2\times a\times 2\\ \Rightarrow a=-500\quad m/{ s }^{ 2 }\\ v=u+at\\ \Rightarrow t=\dfrac { \sqrt { 2000 } }{ 500 } =0.089s \sim 0.09s$

Which of the following statements are true for a moving body?

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If its speed changes, its velocity must change and it must have some acceleration.
0%
If its velocity changes, its speed must change and it must have some directions.
0%
If its velocity changes, its speed may or may not change, and it must have some acceleration.
0%
If its speed changes but direction of motion does not change, its velocity may remain constant.

Explanation

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

Speed is the magnitude of velocity.
There can be a case where there is change in direction but not in magnitude.

$t=0$ , an arrow is fired vertically upwards with a speed of

$98m{ s }^{ -1 }$ . A second arrow is fired vertically upwards with the same speed at

$t=5s$ . Then, (Take

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

0%
the arrows will be at the same height above the ground at $t=12.5s$ .
0%
the two arrows will reach back at their starting points at $t=20s$ and $t=25s$ .
0%
the ratio of the speeds of the first and the second arrows at $t=20s$ will be $2:1$ .
0%
all are correct.

Explanation

$s=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }\\ \Rightarrow 98t-\dfrac { 1 }{ 2 } 9.8{ t }^{ 2 }=98(t-5)-\frac { 1 }{ 2 } 9.8{ (t-5) }^{ 2 }\\ \Rightarrow { t }^{ 2 }-{ (t-5) }^{ 2 }-100=0\\ \Rightarrow 10t=125\\ t=12.5s$
Now, total time for first arrow will be

$s=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }\\ \Rightarrow 98t-\dfrac { 1 }{ 2 } 9.8{ t }^{ 2 }=0\\ t(t-20)=0\\ { t }_{ 1 }=20s\\ \Rightarrow { t }_{ 2 }=25s$
Now at

$t=20s$

$v=u+at\\ { v }_{ 1 }=98\quad \& \quad { v }_{ 2 }=98\dfrac { 1 }{ 2 } \\ \dfrac { { v }_{ 1 } }{ { v }_{ 2 } } =\dfrac { 2 }{ 1 } \\$

A stone is dropped into a well in which the level of water is

$h$ below the top of the well. If

$v$ is velocity of sound, the time

$T$ after which the splash is heard is given by

0%
$T=\displaystyle\frac { 2h }{ v }$
0%
$T=\sqrt { \left( \displaystyle\frac { 2h }{ g } \right) } +\displaystyle\frac { h }{ v }$
0%
$T=\sqrt { \left( \displaystyle\frac { 2h }{ v } \right) } +\displaystyle\frac { h }{ g }$
0%
$T=\sqrt { \left( \displaystyle\frac { h }{ 2g } \right) } +\displaystyle\frac { 2h }{ v }$

Explanation

Time taken by the stone to reach the water level,
By second kinematic equation,

$S = ut + \dfrac{1}{2}gt^{2} \Rightarrow t_{1} = \sqrt{\dfrac{2h}{g}} -------\left ( i \right )$
Time taken by sound to come to the mouth of the well,

$Time = \dfrac{Distance}{Speed} \Rightarrow t_{2} = \dfrac{h}{v} -------\left ( ii \right )$

$\therefore$ Total time

$t_{1} + t_{2} = \sqrt{\dfrac{2h}{g}} + \dfrac{h}{v}$

Figure shows the

$V-T$ graph for two particles

$P$ and

$Q$ . The relative velocity of

$P$ w.r.t.

$Q$ is:

0%
is zero.
0%
is non-zero but constant
0%
continuously decreases
0%
continuously increases

A balloon starts rising from the ground with an acceleration of

$1.25 { m }/{ { s }^{ 2 } }$ . After

$8 \; s$ , a stone is released from the balloon. The stone will (Taking

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

0%
begin to move down after being released
0%
reach the ground in $4$ $s$
0%
cover a distance of $40$ $m$ in reaching the ground
0%
will have a displacement of $50$ $m$

Explanation

$v=1.25*8{ m }/{ s }=10{ m }/{ s }$

$s=\frac { 1 }{ 2 } *1.25*8*8{ m }/{ s=40m }$

now,

$40=-10t+\dfrac { 1 }{ 2 } *10*{ t }^{ 2 }$
or,

$5{ t }^{ 2 }-10t-40=0$
or,

${ t }^{ 2 }-2t-8=0$

hence,

$t=4s$

So B is correct.
s

$=$ 40m.

displacement

$=$ 40m.

Just after being released the stone has an upward velocity, so it will move upwards first. So A is wrong.
distance in an upward direction before stopping d.

$d =\dfrac { { v }^{ 2 }-{ u }^{ 2 } }{ 2g } =\dfrac { { 10 }^{ 2 }-{ 0 }^{ 2 } }{ 2*10 } =5m$

and distance

$=$ s+2d

$=$ 50m
So the distance covered is 50 m and the displacement is 40 m. So C and D are wrong.

A body travels

$2\ m$ in the first two second and

$2.20\ m$ in the next

$4$ second with uniform deceleration. The velocity of the body at the end of

$9$ second is

0%
$-10\ { m }/{ s }$
0%
$-0.20\ { m }/{ s }$
0%
$-0.40\ { m }/{ s }$
0%
$-0.80\ { m }/{ s }$

Explanation

A to B

$2=u\times 2+\displaystyle\dfrac { 1 }{ 2 } \times a\times 2\times 2\Rightarrow 1=u+a$
A to C

$4.20=u\times 6+\displaystyle\dfrac { 1 }{ 2 } \times a\times 6\times 6\Rightarrow 0.7=u+3a$

Let the Body travel from

$A$ to

$B$ in

$2\ s$ for a distance of

$2\ m$
Let the Body travel from

$B$ to

$C$ for next

$4 s$ for a distance of

$2.20\ m$
Velocity after

$9\ s = ?$
For

$A to B$
WKT,

$S = ut + \dfrac{1}{2}at^{2} \Rightarrow 2 = 2u + \dfrac{1}{2}a*2*2 \Rightarrow 1 = u + a ... (1)$
For

$B$ to

$C$
WKT,

$S = ut + \dfrac{1}{2}at^{2} \Rightarrow 4.20 = 6u + \dfrac{1}{2}a*6*6 \Rightarrow 0.7 = u + 3a ... ( 2)$
From

$(1)$ and

$( 2) ,$ we get,

$2a = - 0.3 \Rightarrow a = -0.15 , -ve$ as it is decreasing acceleration .

$u = 1 - a \Rightarrow u = 1 + 0.15 = 1.15$
Now, velocity at

$t = 9 s = v = u + at \Rightarrow v = 1.15 - 0.15*9 = 1.15 - 1.35 = - 0.2 m/s$
velocity is negative as it is decreasing.

A stone is just released from the window of a moving train moving along a horizontal straight track .When observed by a person on the ground,the stone will hit the ground following a

0%
straight line path
0%
circular path
0%
parabolic path
0%
hyperbolic path

A bomb is released from a horizontal flying aeroplane .The trajectory of bomb as observed from ground is

0%
a parabola
0%
a straight line
0%
a circle
0%
a hyperbola

A ball is thrown vertically upwards. It returns

$6 s$ later. Calculate:

$(i)$ the greatest height reached by the ball, and

$(ii)$ the initial velocity of the ball. (Take

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

0%
(i) $40$ $m$ , (ii) $30$ $m$ ${s}^{-1}$
0%
(i) $45$ $m$ , (ii) $30$ $m$ ${s}^{-1}$
0%
(i) $45$ $m$ , (ii) $60$ $m$ ${s}^{-1}$
0%
(i) $45$ $m$ , (ii) $20$ $m$ ${s}^{-1}$

Explanation

The ball is thrown up and it returns in $6 sec$ ,

Time of accent = Time of decent

So, time to reach the highest point $=6/2 = 3 s$

By 2nd equation of motion

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

To calculate height, consider motion of the ball from highest point to the ground

$H = \dfrac{1}{2}gt^{2} = \dfrac{1}{2} \times 10 \times 3^{2} =45\ m$

To calculate projection velocity, consider motion of the ball from projection to return to the point of projection.

$0=ut-\dfrac{1}{2}gt^2$

$u=10 \times 6/2=30\ m/s$

A ball is thrown from rear end of the compartment of train to the front end which is moving at a constant horizontal velocity. An observer

$A$ sitting in compartment and another observer

$B$ standing on the ground draw the trajectory. They will have

0%
equal horizontal and equal vertical ranges.
0%
equal vertical ranges but different horizontal ranges.
0%
different vertical ranges but equal horizontal ranges.
0%
different vertical ranges and different horizontal ranges.

Explanation

In vertical direction, ball has zero initial velocity but same value of

$g$ w.r.t observer A & B,so they will draw trajectory of equal vertical range
But along horizontal direction, for observer

$A$ , ball is moving with velocity at which it is thrown w.r.t. train while for

$B$ , it is moving with a velocity equal to train

$+$ velocity at which it is thrown w.r.t groundtrain

A rocket is fired upward from the earth's surface such that it creates an acceleration of

$19.6 { m }{ { s }^{- 2 } }$ . If after

$5 s$ , its engine is switched off, the maximum height of the rocket from earth's surface would be

0%
$980\ m$
0%
$735\ m$
0%
$490\ m$
0%
$245\ m$

Explanation

concept are attached, rest calculation is:
Velocity when the engine is switched off

$v=19.6\quad *\quad 5\quad =\quad 9.8\quad { m }/{ s }$

${ h }_{ max }={ h }_{ 1 }+{ h }_{ 2 }$
where,

${ h }_{ 1 }=\dfrac { 1 }{ 2 } a{ t }^{ 2\quad }\& \quad { h }_{ 2 }=\dfrac { { v }^{ 2 } }{ 2a }$

${ h }_{ max }=\dfrac { 1 }{ 2 } *19.6*5*5+\dfrac { 98*98 }{ 2*9.8 }$

$= 245+490=735 m$

$B$ is correct answer

From a

$200 m$ high tower, one ball is thrown upwards with speed of

$10 { m }/{ s }$ and another is thrown vertically downwards at the same speed simultaneously. The time difference of their reaching the ground will be nearest to

0%
$12 s$
0%
$6 s$
0%
$2 s$
0%
$1 s$

Explanation

The ball thrown upward will have zero velocity in

$1\: s$ . It returns back to thrown point in another

$1\: s$ with the same velocity as second. Thus the difference will be

$2\: s$ .

A car starting from rest accelerates uniformly to acquire a speed

$20\ kmh^{-1}$ in 30 min. The distance travelled by car in this time interval will be

0%
600 km
0%
5 km
0%
6 km
0%
10 km

Explanation

Here,

$u = 0, v = 20 km/h, t = 30 min = 0.5 h$

Putting these values in

$v=u+at$

$20 = 0+a*0.5$
We get,

$a = 40 km/h^2$

Putting the values in

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

$S = 0*0.5+\dfrac{1}{2}*40*0.5^2$

$S = 5 km$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

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 position(S) and velocity(v) of the particle at

$t = 5$ s

0%
$S = 25 m, v = 0 m/s$
0%
$S = -25 m, v = 0 m/s$
0%
$S = 5 m, v = 10 m/s$
0%
$S = 0 m, v = 0 m/s$

Explanation

Displacement at

$t =5 s$ is

$S= u \, t +\dfrac{1}{2} at ^2$

$=10 \times 5 + \dfrac{1}{2} \times (-2.0) \times (5)^2$

$=50-25=25 \, m$

i.e., after

$5 s$ , the particle will be at distance

$25 m$ from the starting point.

Velocity at

$t = 5 s$ is

$v = u + at$

$v = 10+(-2)\times 5=0$

A body moves from rest with uniform acceleration and travels 270 m in 3 s. Find the velocity of the body at 10 s after the start.

0%
$600\, m\, s^{-1}$
0%
$6\, m\, s^{-1}$
0%
$60\, m\, s^{-1}$
0%
$100\, m\, s^{-1}$

Explanation

$u=0, s=270 m, t=3 s$

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

$270 = 0\times 3+\dfrac{1}{2}\times a\times 3^2$

$a= 60\ m/s^2$

Now putting the value of

$a$ in

$v=u+at$ ,
Velocity after

$t=10 s$ is given by

$v=0+60\times 10 = 600\ m/s$

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 initial velocity.

0%
$15 \,m \,s^{-1}$
0%
$10 \,m \,s^{-1}$
0%
$30 \,m \,s^{-1}$
0%
$0 \,m \,s^{-1}$

Explanation

For uniformly accelerated motion,

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

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

$u=15\ m/s$

(a) How long will a stone take to fall to the ground from the top of a building

$80 m$ high and (b) what will be the velocity of the stone on reaching the ground? (Take

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

0%
(a) $14$ $s$ , (b) $30$ $m$ ${s}^{-1}$
0%
(a) $4$ $s$ , (b) $30$ $m$ ${s}^{-1}$
0%
(a) $10$ $s$ , (b) $40$ $m$ ${s}^{-1}$
0%
(a) $4$ $s$ , (b) $40$ $m$ ${s}^{-1}$

Explanation

a.)

$s=ut+0.5at^2$

$80=0+0.5(10)t^2$

$t=4\ s$
b.)

$v=u+at$

$=10\times 4=40\ m/s$

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

$19.6 m {s}^{-1}$ . The ball reaches the ground after

$5 s$ . Calculate :

$(i)$ the height of the tower,

$(ii)$ the velocity of ball on reaching the ground. Take

$g=9.8 m {s}^{-2}$ .

0%
$(i) 24.5 m$ , $(ii) 29.4$ $m$ ${s}^{-1}$
0%
$(i) 24.5 m$ , $(ii) 19.4$ $m$ ${s}^{-1}$
0%
$(i) 24.5 m$ , $(ii) 28$ $m$ ${s}^{-1}$
0%
$(i) 25 m$ , $(ii) 29.4$ $m$ ${s}^{-1}$

Explanation

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

$=19.6 \times 5 - \dfrac{1}{2} \times 9.8 \times 5^2$

$=-24.5\ m$

Hence, height of tower

$= 24.5 m$

$v=u+at$

$=19.6-9.8\times 5=-29.4\ m/s$

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 ball is thrown vertically upwards with an initial velocity of

$49 m {s}^{-1}$ . Calculate: (i) the maximum height attained, (ii) the time taken by it before it reaches the ground again. (Take

$g=9.8 m {s}^{-2}$ )

0%
(i) 122.5 m, (ii) 10 s
0%
(i) 120.5 m, (ii) 10 s
0%
(i) 122.5 m, (ii) 20 s
0%
(i) 120 m, (ii) 10 s

Explanation

(i)

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

$0=49^2-2(9.8)(s)$

$s=122.5\ m/s$

(ii)

$v=u+at$

$0=49-9.8t$

$t=5\ s$
total time

$=10\ s$

A boy on a cliff

$49 m$ high drops a stone.One second later,he throws a second after the first .They both hit the ground at the same time.With what speed did he throw the second stone?

0%
$12.1 m/s$
0%
$10 m/s$
0%
$21.1 m/s$
0%
$15.1 m/s$

Explanation

If time for first stone is

$t$ , time for second stone is

$t-1$

$s=49m$ is the same for both stones,

$u=0$ for first stone

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

$v^2=0+2(9.8)(49)$

$v=31 m/s$

$v=u+at$

$31=0+9.8t$

$t=3.16 s$

$t'=2.16 s$

$s=ut+0.5at^2$

$49=u'(2.16)+0.5(9.8)(2.16)^2$

$u'=12.1 m/s$

A stone is thrown vertically upward with an initial velocity of

${40 ms^{-1}}$ ,Taking g=

${10 ms^{-2}}$ , find the maximum height reached by the stone.

0%
$100 m$
0%
$60 m$
0%
$80 m$
0%
$120 m$

Explanation

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

$0=40^2+2(-10)s$

$s=\dfrac{40\times40}{2\times10}$

$s=80 m$

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,

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

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

A ball is thrown vertically upwards with a velocity of

${20 ms^{-1}}$ from the top of a multi storey building.The height of the point where the ball is thrown 25 m from the ground.How long will it be before the ball hits the ground ? Take

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

0%
t=5s
0%
t=10s
0%
t=15s
0%
t=20s

Explanation

Time to reach maximum height can be obtained from

$v=u+at$

$0=20+(-10)t$

$t=2 s$

$s=ut+0.5at^2=20(2)+0.5(-10)(2)^2= 20 m$
Thus, total distance for maximum height is 45 m

$s=ut+0.5at^2$

$45=0+0.5(10)(t')^2$

$t'=3 s$
Total time= 3+2= 5s

A stone is thrown vertically upwards. When stone is at a height half of its maximum height, its speed is

$\displaystyle 10 \ m/s;$ then the maximum height attained by the stone is (g=

$\displaystyle 10 \ m/s^{2}$ )

0%
$25\ m$
0%
$10\ m$
0%
$15\ m$
0%
$20\ m$

Explanation

Let maximum height be H. At H, v=0.

Thus,

$0-{u}^{2}=-2gH$

or,

$u=\sqrt{2gH}$

Now, at H/2, v=10.

Thus,

${10}^{2}-{u}^{2}=-2g(H/2)$

or,

$100-2gH=-gH$

giving,

$H=\frac{100}{10}=10m$ (taking g=10)

A car starts from rest and uniformly accelerates in a straight line. In the first second the car covers a distance of

$2$ m. The velocity of the car at the end of

$2^{nd}$

$second$ will be

0%
$\displaystyle 4.0\ ms^{-1}$
0%
$\displaystyle 8.0\ ms^{-1}$
0%
$\displaystyle 6\ ms^{-1}$
0%
None of these

Explanation

distance covered by car in t seconds

$=0*t+\frac { 1 }{ 2 } a{ t }^{ 2 }$
for

$t=1$ distance is 2m.

$2 = \frac { 1 }{ 2 } a*{ 1 }^{ 2 }\\ a=4 m/s^2$
velocity after 2 seconds.

$v= u +at\\=0+4*2 = 8m/s$

A stone is allowed to fall from the top of a tower

$100 m$ high and at the same time another stone is projected vertically upwards from the ground with a velocity of

${25 ms ^{-1}}$ .Calculate where the two stones will meet.(Take

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

0%
10 m
0%
40 m
0%
20 m
0%
50 m

Explanation

For stone moving downward,acceleration due to gravity ,

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

Initial velocity (u)

$=$ o;distance, s

$=$ 100-x;time, t

$=$ ?

${s=ut+\dfrac{1}{2}gt^2}$ , or (100-x)=

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

${\Rightarrow 100 -x=5t^2}$ .....(1)

For the stone moving vertically upwards,

For the stone moving vertically upwards,initial velocity,

${u=25 ms ^{-1}}$ ;time (t)=?;

acceleration due to gravity

${g=-10 ms^{-1}}$

[In upward direction ,g is taken -ve]

Distance,

$s=x$

we know

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

${x=25\times t +\dfrac{1}{2}\times (-10 t^2)}$

${\Rightarrow x=25 t-5t^2}$ .....(2)

Substituting the value of x from (2) in (1) we get,

${100-(25t-5t^2)=5t^2}$

${100-25 t+5t^2=5t^2}$

$25 t=100$ or

$t=4s$

Put the value of tin (1)

${\Rightarrow 100 -x =5(4)^2}$

${\Rightarrow 100-x=80}$ or x

$=$ 20 m

${\therefore}$ The stones will meet at a height of 20m

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

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

If a car at rest accelerates uniformly to a speed of

$144\ km/h$ in

$20$ second, it covres a distance of :-

0%
$20\ m$
0%
$400\ m$
0%
$1440\ m$
0%
$2980\ m$

Explanation

speed of car after 20 sec.

$=144\quad km/h\quad =\quad 144*\frac { 5 }{ 18 } \quad m/s=40m/s$

$v=u+at\\40=0+20a\\a=2m/s^2$
Distance covered

$s=ut+1/2 at^2\\=0*20+\frac 12 *2*{20}^2\\=400 m$

A ball thrown up vertically returns to the thrower after 6 s Find it's initial velocity.

0%
20 m/s
0%
40 m/s
0%
35 m/s
0%
30 m/s

Explanation

The ball returns to the thrower in

$6$ s,thus the time for its upward journey

$=$

${6/2}$

$=$ 3s
For the upward motion of ball.
Initial velocity

$u=$ ?;final velocity

$v=$ 0
(

${\because }$ Ball comes to rest)
Time ,

$t=3$ s
Acceleration due to gravity ,

${g=-10 ms^{-2}}$
[In upward direction ,

$g$ is taken -ve]

$v$

$=$

$u+gt$
or

${0=u-10 \times 3,or -u=-30}$
or

${u=30 ms^{-1}}$

${\therefore}$ Initial velocity of ball is

${30 ms ^{-1}}$

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

$A$

A body starts from rest is moving under a constant acceleration up to 20 s . If it moves

$S_{1}$ Distance in first 10 s, and

$S_{2}$ distance in next 10 s then

$S_{2}$ will be equal to:

0%
$S_{1}$
0%
$2S_{1}$
0%
$3S_{1}$
0%
$4S_{1}$

Explanation

$\displaystyle s_{1}=0+\frac{1}{2}a(10)^{2}=50a.............(i)$
after

$10s\;\;\;\;\;\;\;v=10a.....(ii)$
Now

$\displaystyle s_{2}= (10a)t+\frac{1}{2}a(10)^{2}$

$\displaystyle s_{2}= (10a)10+\frac{1}{2}a(100)$

$s_{2}=100a +50a=150a=3(50a)\Rightarrow s_{2}=3s_{1}$

An aeroplane is flying in a horizontal direction at

$600 km/hr$ at a height of

$6 kms$ and is advancing towards a point which is exactly over a target on earth. At that instant the pilot releases a ball which on descending the earth strike the target. The falling ball appears-

0%
To the pilot in the aeroplane, as falling along a parabolic path
0%
To a person standing near the target, as falling exactly vertical
0%
To a person standing near the target, as describing a parabolic path
0%
To the pilot sitting in the aeroplane, as falling in a zigzag path

A particle travels

$10\ m$ in first

$5\ s$ ,

$10\ m$ in next

$3\ s$ . Assuming constant acceleration, what is the distance traveled in the next two seconds.

0%
$\displaystyle \frac{20}{3}\ m$
0%
$20\ m$
0%
$60\ m$
0%
$180\ m$

Explanation

In the first

$5$ sec:

$s = ut+0.5at^2$

$10 = 5u + 0.5a \times 25$

In the next

$3$ sec:

$20 = 8u + 0.5a \times 64$

Solving above two equations will give us:

$u = \dfrac{7}{6} , a = \dfrac{1}{3}$

Therefore, in next

$2$ sec distance traveled is:

$S = \dfrac{7}{6} \times 10 + 0.5 \times \dfrac{1}{3} \times 100 = \dfrac{170}{6}$

Therefore distance travelled in next

$2$ sec is

$=\dfrac{170}{6}-20 = \dfrac{20}{3}$

A body travels

$200\ cm$ in the first two seconds and

$220\ cm$ in the next

$4\ s$ with same acceleration.The velocity of the body at the end of the

$7^{th}$ second is

0%
$5\ cm/s$
0%
$10\ cm/s$
0%
$15\ cm/s$
0%
$20\ cm/s$

Explanation

$200=u\times 2+\dfrac{1}{2}a(2)^{2}$

$100=u+a..........(i)$

$420=u(6)+\dfrac{1}{2}a(6)^{2}$

$70=u+3a..........(ii)$

from $(i)$ and $(ii)$ $2a=-30\Rightarrow a=-15cm/s^{2}$

$\therefore u=100-a=100-(-15)=115\;cm/s$

$u_{7^{th}}=u+at=115+(-15)(7)=115-105=10\;cm/sec$

The distance between them at time

$t$ is:-

0%
$\;\sqrt{(200)^2+(100)^2}\:m$
0%
$\;\sqrt{(200-4t)^2+(100-2t)^2}\:m$
0%
$\;[(200-4\,t)+(100-2\,t)]\:m$
0%
$\;\sqrt{(200-2\,t)^2+(100-4\,t)^2}\:m$

Explanation

Given :

$V_V = 2 m/s$

$V_W =4 m/s$

$\therefore$ Distance traveled by ship V in time

$t$ ,

$l_1 = 2t$

Also distance traveled by ship W in time

$t$ ,

$l_2 = 4t$

$\implies$

$OA = 100 - 2t$ and

$OB = 200 -4t$

$\therefore$ Distance between them at time

$t$ ,

$AB = L =\sqrt{OA^2 + OB^2}$

$\implies$

$L = \sqrt{(100-2t)^2 + (200 -4t)^2}$ m

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

0%
Momentum
0%
Displacement
0%
Acceleration
0%
Impulse

Explanation

Acceleration is the rate of change of the velocity of an object with respect to time. Accelerations are vector quantities. The orientation of an object's acceleration is given by the orientation of the net force acting on that object.

$\text{acceleration} = \dfrac{\text{Change in the velociity}}{\text{time}}$

CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 4 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 4 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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