MCQ Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 8

A car travels with a uniform velocity of

$20\,\,m{s^{ - 1}}$ for 5 sec. The brakes are then applied and the car is uniformly retarded. It comes to rest in further 8 sec by drawing a graph between velocity and time-find.

0%
The distance traveled in the first 5 sec.
0%
The distance traveled after the brakes are applied.
0%
Total distance traveled.
0%
Acceleration during the first 5 sec.
0%
Acceleration during the last 8 sec.

An aircraft is flying at a height of

$3400 \,m$ above the ground. If the angle subtended at a ground observation point by the aircraft positions

$10 \,s$ apart is

$30^o$ , then the speed of the aircraft is:

0%
10.8 $ms^{-1}$
0%
1963 $ms^{-1}$
0%
108 $ms^{-1}$
0%
196.3 $ms^{-1}$

Explanation

O is the observation point at the ground. A and B are the positions of aircraft for which

$\angle$ AOB = 30

$^{\circ}$ .Time taken by aircraft from A to B is 10 s.

$\Delta$ AOB,

$tan 30^{\circ}=\dfrac{AB}{3400}; \ \ \ AB= 3400 \ tan 30^{\circ}=\dfrac{3400}{\sqrt{3}}$ m

$\therefore$ Speed of aircraft,

$v = \dfrac{AB}{10}=\dfrac{3400}{10\sqrt{3}} = 196.3 ms^{-1}$

1050334_940484_ans_e09c19fe8fef4a74994e24a56142a311.JPG

A ball rolls off the top of a stairway horizontally with a velocity of

${ 4.5\ ms }^{ -1 }$ . Each steps is

$0.2\ m$ high and

$0.3\ m$ wide. If

$g$ is

${ 10\ ms }^{ -2 }$ , and the ball strikes the edge of the

$nth$ step, then

$n$ is equal to:

0%
$9$
0%
$10$
0%
$11$
0%
$12$

Explanation

It strikes on

${ n }^{ th }$ step edge.

Therefore,

$4.5t=0.3n$

$\Rightarrow \quad t=\dfrac { n }{ 15 }$

$\dfrac { 1 }{ 2 } { gt }^{ 2 }=0.2n$

$\dfrac { 1 }{ 2 } \times 10\times \dfrac { { n }^{ 2 } }{ { 15 }^{ 2 } } =0.2n$

$\Rightarrow \quad n=9$

1086797_980813_ans_55fc10ad1fdd43a29778f8a6ec9767d7.PNG

A man can swim with a speed of 4

$km h^{-1}$ in still water. He crosses a river 1 km wide that flows steadily at 3

$kmh^{-1}$ . If he makes his strokes normal to the river current, how far down the river does he go when he reaches the other bank?

0%
500 m
0%
600 m
0%
750 m
0%
850 m

Explanation

Time to cross the river,

$t=\dfrac{Width \ of \ river}{Speed \ of \ man}+\dfrac{1km}{4 kmh^{-1}}$

$=\dfrac{1}{4}h$
Distance moved along the river in time t,

$s= v_{r} \times t = 3kmh^{-1}\times \dfrac{1}{4}h$

$= \dfrac{3}{4}km = 750m$

1069240_940482_ans_e0e350f6431542939fcac78e8780b11c.PNG

The displacement of a body at any time

$t$ after starting is given by

$s = 15t - 0.4 t^{2}$ . Find the time when the velocity of the body will be

$7\ ms^{-1}$ .

0%
$t=20\,s$
0%
$t=10\,s$
0%
$t=30\,s$
0%
$t=40\,s$

Explanation

$S=15t-0.4t^{2}$

$\dfrac{ds}{dt}=15-0.8t$

$\vartheta =\dfrac{ds}{dt}$ so,

$\vartheta =15-0.8t$

Given,

$\vartheta =7$

$7=15-0.8t$

$t=10$ sec

A balloon is ascending vertically with an acceleration of 1

$ms^{-2}$ . Two stones are dropped from it at an interval of 2 s. Find the distance between them 1.5 s after the second stone is released.

0%
35 m
0%
45 m
0%
55 m
0%
None of these

Explanation

(This method can be understood in a proper way after studying relative velocity.)
If we work from the frame of the balloon, then the acceleration of each stone w.r.t. the balloon will be g + a after releasing from it. The initial velocity of each stone will be zero w.r.t. balloon.

$S_1 = 1/2 (g + a)(3.5)^2$ ,

$S_2 = (1/2)(g + a)(1.5)^2; x = S_1 - S_2$ = 55m

A particle is moving in

$xy$ -plane with

$y = x/2$ and

$v_{x} = 4 - 2t$ . Choose the correct options.

0%
Initial velocities in x and y directions are negative
0%
Initial velocities in x and y directions are positive
0%
Motion is first retarded, then accelerated
0%
Motion is first accelerated, then retarded

Explanation

$y = \dfrac {x}{2}$ implies that the particle is moving in a straight line passing through the origin.

$v_{x} = 4 - 2t, v_{x} = u_{x} + a_{x}t, u_{x} = 4, a_{x} = -2$
Now,

$y = \dfrac {x}{2}\Rightarrow \dfrac {dy}{dt} = \dfrac {1}{2} \dfrac {dx}{dt}$

$v_{y} = \dfrac {1}{2} v_{x} = 2 - t, v_{y} = u_{y} + a_{y}t$ ,

$u_{y} = 2$ and

$a_{y} = -1$ .
1554326_986259_ans_9b80458f628c4000a653d8965467ea49.jpg

Two cities

$A$ and

$B$ are connected by a regular bus services with buses plying in either direction every

$T$ seconds. The speed of each bus is uniform and equal to

$V_{b}$ . A cyclist cycles from

$A$ to

$B$ with a uniform speed of

$V_{c}$ . A bus goes past the cyclist in

$T_{1}$ second in the direction

$A$ to

$B$ and every

$T_{2}$ second in the direction

$B$ to

$A$ . Then

0%
$T_{1} = \dfrac {V_{b} T}{V_{b} + V_{c}}$
0%
$T_{2} = \dfrac{V_{b} T}{V_{b} - V_{c}}$
0%
$T_{1} = \dfrac {V_{b} T}{V_{b} - V_{c}}$
0%
$T_{2} = \dfrac {V_{b} T}{V_{b} + V_{c}}$

Explanation

Distance between two buses on road is

$V_bT$ .

For

$A$ to

$B$ direction :

Distance = Relative velocity

$\times$ Time

$V_bT=(V_b-V_c)T_1\Rightarrow T_1=\dfrac{V_bT}{V_b-V_c}$

For

$B$ to

$A$ Direction :

$V_bT=(V_b+V_c)T_2 \Rightarrow T_2=\dfrac{V_bT}{V_b+V_c}$

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%
$8 s$

Explanation

$\textbf{Hint}$ Use equation of motion.
2149240_986177_ans_2cf10cdfaebe451fb1bad6427028cbc5.jpeg

A body is dropped from a balloon moving up with a velocity of

$4\ m/s$ when the balloon is at a height of

$120.5\ m$ from the ground. The height of the body after

$5\ s$ from the ground is?

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

0%
$8\ m$
0%
$12\ m$
0%
$18\ m$
0%
$24\ m$

Explanation

Given,

$u=4m/s$

$g=9.8m/s^2$

$t=5sec$

$H=120.5m$

From 2nd equation of motion,

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

$-S=4\times 5-\dfrac{1}{2}\times 9.8\times 5\times 5$

$S=-20+122.5=102.5m$

Height of the body after 5 sec,

$h=H-S$

$h=120.5-102.5$

$h=18m$

The correct option is C.

Newtons law are not valid in

0%
Both inertial as well as non- inertial frame of reference.
0%
A frame moving with constant velocity w.r.t. an inertial frame.
0%
All reference frames which are at rest w.r.t. an inertial frame.
0%
The reference frame attached to the earth.

Explanation

Newton's Law is only valid in inertial frame of reference. If there is a non - inertial frame of reference then it won't be valid.

Hence option

$\textbf A$ is the correct answer.

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

$f/2$ to come to rest. If the total distance traversed is

$5\ S$ , then

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

Explanation

$s_{1} + s_{2} + s_{3} = 5s$

$\Rightarrow s + vt + \dfrac {v^{2}}{f} = 5s$

$\Rightarrow 4s = vt + \dfrac {v^{2}}{f} = \sqrt {2fs}\times t + \dfrac {2fs}{f}$

$\Rightarrow 2s = t\sqrt {2fs} \Rightarrow s = \dfrac {1}{2}ft^{2}$ .
930069_1012174_ans_bb6046a0828c415c9cdde20371bf0003.jpg

A ball of mass

$0.2kg$ is thrown vertically upwards by applying a force by hand. If the hand moves

$0.2m$ which applying the force and the ball goes upto

$2m$ height further, find the magnitude of the force. Consider

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

0%
$22N$
0%
$4N$
0%
$16N$
0%
$20N$

Explanation

(i)

${ v }^{ 2 }={ u }^{ 2 }+2ay$

$0={ v }^{ 2 }-2(g)2$
(ii)

${ v }^{ 2 }=0+2a(0.2)$

$a=100$
(iii)

$F=ma=0.2\times 100=20N$

A body moves for a total of nine second starting from rest with uniform acceleration and then with uniform retardation, which is twice the value of acceleration and then stops. The duration of uniform acceleration

0%
$3\ s$
0%
$4.5\ s$
0%
$5\ s$
0%
$6\ s$

Explanation

Consider the acceleration of the is

$\alpha$ and retardation is

$\beta$

and according to question retardation is twice that acceleration then

$\beta = 2a$

Body moves with acceleration

$\alpha$ for

${ t_{ 1 } }$ sec

$v=u+at$

$v=0+\alpha { t_{ 1 } }$

$.....(1)$

Body retards with

$2\alpha$ for time

${ t_{ 2 } }$ sec and then body becomes rest .

$v=u+at$

Here,

$\begin{array}{l} u=\alpha { t_{ 1 } } \\ v=0 \\ a=-2\alpha \end{array}$

$\begin{array}{l} 0=\alpha { t_{ 1 } }-2\alpha { t_{ 2 } } \\ { t_{ 1 } }=2{ t_{ 2 } } \end{array}$

${t_1} + {t_2} = 9\sec$

Then,

${t_1} = 3\sec$

${t_2} = 6\sec$

Therefore,

$3\;sec$ body moves with uniform acceleration.

Hence, Option

$A$ is the correct answer.

A car starts from rest and acceleration at

$4\ {m/s}^{2}$ for

$5\ s$ . After that it moves with constant velocity for

$25\ s$ and then retards at

$2\ m/s^{2}$ until it comes to rest. The total distance travelled by the car is

0%
$650\ m$
0%
$527\ m$
0%
$675\ m$
0%
$573\ m$

Explanation

For

$0-5 sec$ ,

$a= 4m/s^2$ , Let velocity at

$5 sec$ is

$v$ and distance is

$S_1$

Using

$v=u+at$

$\Rightarrow v= 0+4\times 5=20 m/s$

Using

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

$\Rightarrow S_1=0+\dfrac{1}{2}\times 4\times 5^2=50m$

For next

$25sec$

$v= 20 m/s$

$\Rightarrow S_2= 20\times 25= 500 m$

For comes to rest,

$u= 20 m/s$ ,

$a=-2 m/s^2$

Using

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

$\Rightarrow 0=20^2-2\times 2\times S_3$

$\Rightarrow S_3= 100 m$

Total distance

$= S_1+S_2+S_3= 50+500+100=650 m$

A parachutist after bailing out falls

$50\ m$ without friction. When parachute opens, it decelerates at

$2\ m/s^{2}$ . He reaches the ground with a speed of

$3\ m/s$ . At what height, did he bail out?

0%
$91\ m$
0%
$182\ m$
0%
$293\ m$
0%
$111\ m$

Explanation

$A\rightarrow B$

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

$v^{2} = 0 +2\times 9.8\times 50$

$v = \sqrt {980} m/s$
From

$B\rightarrow C$

$V^{2} = u^{2} + 2as$

$(3)^{2} = 980 - 2\times 2\times s$

$s = 242.75$
Total height

$= 242.72 + 50 =292.75\approx 293\ m$ .
930076_1012176_ans_bfe4d942d3484c21a4b4239d575ee5e7.png

A bullet of mass m looses

$\left(\dfrac{1}{n}\right)^2$ of its velocity passing through one plank. The number of such planks that are required to stop the bullet can be:

0%
$\dfrac{n^2}{2n-1}$
0%
$\dfrac{2n^2}{n-1}$
0%
Infinite
0%
n

Explanation

Let uniform resistance of 'a' acts on bullet as it passes through plank. If 'd' is thickness one plank

$\therefore \dfrac{v^2(n-1)^2}{n^2}=v^2- 2ad$

$\Rightarrow a= v^2\left[1 \dfrac{(1-n)^2}{n^2}\right] =\dfrac{(2n-1)}{2dn^2}v^2$
If N is plank are arranged side by side, then

$v^2= 2a(Nd)$

$\Rightarrow N=\dfrac{v^2}{2ad}=\dfrac{v^22dn^2}{2d\times (2n-1)v^2}=\dfrac{n^2}{(2n-1)}$

An object starts 5m from origin and moves with an initial velocity of 5

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

$m{s^{ - 2}}$ . After 10 sec, the object is how far from the origin?

0%
150 m
0%
145 m
0%
155 m
0%
55 m

Explanation

Displacement in

$10seconds$ is

$ut +at^2/2$

$=5\times10 +2\times 10^2/2=150meter$

Final position is initial position+displacement

$5+150=155meter$

A train starts from rest from a station with acceleration

$0.2\ m/{s^2}$ on a straight track and then comes to rest after attaining maximum speed on another station due to retardation

$0.4\ m/{s^2}$ . If total time spent is half an hour, then a distance between two stations is [ Neglect the length of the train]

0%
$216 Km$
0%
$512 Km$
0%
$728 Km$
0%
$1296 Km$

Explanation

Given,

Velocity ${{v}_{1}}$ achieve in time ${{t}_{1}}$ ,due to acceleration, ${{a}_{1}}=0.2\,m{{s}^{-2}}$

${{v}_{1}}={{u}_{1}}+a{{t}_{1}}$

$\Rightarrow {{v}_{1}}=0+0.2{{t}_{1}}$

Velocity reduce to zero in time ${{t}_{2}}$ , due to retardation, ${{a}_{2}}=-0.4\,m{{s}^{-2}}$

${{v}_{2}}={{v}_{1}}+a{{t}_{2}}$

$\Rightarrow 0={{v}_{1}}-0.4{{t}_{2}}$

Add time

${{t}_{1}}+{{t}_{2}}=\dfrac{{{v}_{1}}}{0.2}+\dfrac{{{v}_{1}}}{0.4}$

$\Rightarrow 30\times 60={{v}_{1}}\times 7.5$

$\Rightarrow {{v}_{1}}=240\,m/s$

Apply kinematic equation

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

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

Total distance ${{S}_{1}}+{{S}_{2}}=\dfrac{{{v}_{1}}^{2}}{2{{a}_{1}}}+\dfrac{{{v}_{1}}^{2}}{2{{a}_{2}}}=\dfrac{{{240}^{2}}}{2\times 0.2}+\dfrac{{{240}^{2}}}{2\times 0.4}=216\,km$

Two cars

$A$ and

$B$ are initially

$100\ m$ apart with

$A$ behind

$B$ . Car

$A$ starts from rest with a constant acceleration

$2\ m/s^{2}$ towards car

$B$ and at the same instant car

$B$ starts moving with constant velocity

$10\ m/s$ in the same direction. Time after which car

$A$ overtakes car

$B$ is:

0%
$14.36\ s$
0%
$16.18\ s$
0%
$10\ s$
0%
$16.54\ s$

Explanation

Distance traveled by car

$A= S_A$

Distance traveled by car

$B= S_B$

According to formula

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

$S_A=0t+\frac {1}{2} \times 2t^2$

$S_B=10t+\frac {1}{2} \times 0t^2$

Car

$A$ will overtake car

$B$ when

$S_A=S_B+100$

Putting the value

$S_A$ and

$S_B$

$+2=10t+100$

$+2-10t-100=0$

After solving for

$t$ , we get

$t=16.18\, \sec$ possible

$t=-6.10\, \sec$ not possible

So option B is correct.

A body of mass

$m$ accelerates uniformly from rest to velocity

${v}_{1}$ in time interval

${T}_{1}$ . The instantaneous power delivered to the body as a function of time

$t$ is:

0%
$\dfrac { { mv }_{ 1 }^{ 2 } }{ { T }_{ 1 }^{ 2 } } t$
0%
$\dfrac { { mv }_{ 1 } }{ { T }_{ 1 }^{ 2 } } t$
0%
${ \left( \dfrac { { mv }_{ 1 } }{ { T }_{ 1 } } \right) }^{ 2 }t$
0%
$\dfrac { { mv }_{ 1 }^{ 2 } }{ { T }_{ 1 } } { t }^{ 2 }$

Explanation

Let a be the constant acceleration

Using 1st equation of motion

$v_1=u+aT_1$

$v_1=0+aT_1$

$a=\dfrac{v_1}{T_1}..........(i)$

At any instant t, the velocity v of the body

$v=0+at$

$v=\dfrac{v_1}{T_1}\cdot t.........(ii)$

$\because P=Fv$

$P=mav$

$P=m\left(\dfrac{v_1}{T_1}\right)\left(\dfrac{v_1}{T_1}\cdot t\right)$

$P=\dfrac{mv_1^2}{T_1^2}t$

A ball dropped from the top of the tower falls first half height of the tower in 10s. The total time spent by a ball in the air is $\left[ {Take\;g = 10m/{s^2}} \right]$

0%
$14.14s$
0%
$15.25s$
0%
$12.36s$
0%
$17.36s$

Explanation

Let the height of tower is h. Now considering the equation as

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

$h=ut+\dfrac{1}{2}a{{t}^{2}}...........(1)$

$here,\,\,h=\dfrac{h}{2}$

In 10 sec, object is reaching half the height of tower.

Initially the body is rest so u = 0

It is also given that a = 10 ms^-2

Time to reach first half height of tower t = 10 sec

Put the values in (1)

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

$h=1000\,\,m$

The total time spent by a ball in the air is given by the relation as

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

$1000=\dfrac{1}{2}\times 10\times {{t}^{2}}$

$t=10\sqrt{2}$

$t=14.14\,\sec$

If the time of acceleration is

${t}_{1}$ , then the speed of the car at

$t={t}_{1}$ is:

0%
$2\ {t}_{1}$
0%
$4\ {t}_{1}$
0%
$>2\ {t}_{1}$
0%
$<2\ {t}_{1}$

A body starts from rest with a uniform acceleration of 2m/s

$^2$ for 10 sec , with moves with constant speed for 30 sec then decelerates by 4 m/s

$^2$ to zero . What is the distance covered by the body?

0%
750 m
0%
850 m
0%
950 m
0%
1050 m

Explanation

$u=0$

$a$ =

$2m/S^2$

$t=10 Second$

$S_1$ =

$ut+0.5at^2$

$s_1$ =

$0+0.5\times 2\times 100$ =

$100m$

$S_2$ =

$v+2$

$v=u+at$ =

$0+20$ =

$20m/s$

$s_2$ =

$20\times 30$ =

$600$ m

$0^2-v^2$ =

$2(-a)s_3$

$s_3$ =

$\frac{400}{8}$ =50m

$s=s_1+s_2+s_3$ =

$100+600+50$ =

$750$

A body freely falling from a height

$h$ describes

$\dfrac {7h}{16}$ in the last second of its fall. The height

$h$ is

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

0%
$40\ m$
0%
$45\ m$
0%
$80\ m$
0%
$160\ m$

Explanation

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

u=0

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

distance covered in

$t^{th}$ sec =

$s_t=\dfrac{7g}{16}=u+\dfrac{a}{2}(2t-1)=\dfrac{a}{2}(2t-1)$

solving both the equations we get

$7t^2-32t+16=0$

$t=4secs$

$\dfrac{1}{2}g4^2=80m$

A car travelling with a velocity of

$80 km/h$ slowed down to

$44 km/h$ in

$15 s$ . The retardation is

0%
$0.67m/{s^2}$
0%
$1m/{s^2}$
0%
$1.25m/{s^2}$
0%
$1.5m/{s^2}$

Explanation

$\begin{array}{l}v = u + at\\44 \times \dfrac{5}{{18}}m/s\\ = 80 \times \dfrac{5}{{18}}m/s + a \times 15\\ - 36 \times \dfrac{5}{{18}} = a \times 15\\a = \dfrac{{ - 2}}{3}m/{s^2}\\retardation = 0.67m/{s^2}\end{array}$

A body moving along the positive $x-axis$ with uniform acceleration of $- 4m{s^{ - 2}}$ . Its velocity at $x=0$ is $10m{s^{- 1}}$ . The time taken by the body to reach a point at $x = 12cm$ is:

0%
$(2s)$
0%
$(0.25s)$
0%
$(3s)$
0%
$(0.0125s)$

Explanation

Acceleration

$a=-4 m/s^2$

Initial velocity

$u=10 m/s$

Distance traveled

$x=12 cm=12\times 10^{-2} m$

From kinematics,

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

$v^2=u^2+2ax$

$=100+2\times (-4) \times 12 \times 10^{-2}$

$=100-0.96=99.04$

$v=9.95 m/s$

$v=u+at$

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

$=\dfrac{9.95-10}{-4}=0.0125 s$

A free falling body travels ___ of total distance in 5th second.

0%
8 %
0%
12 %
0%
25 %
0%
36 %

Explanation

The correct option is D

Given,

A free falling body

Distance traveled in 5s

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

$u=0,g=10m/s$

$=\dfrac{1}{2}\times10\times5^2$

$=125m$

Distance travelled in 5^th sis:

Distance travelled in 5s- Distance travelled in 4s

Distance travelled in

$4s=\dfrac{1}{2}\times10\times4^2$

$=80m$

So,

$s_5-s_4=125-80=45$

Therefore,

$\dfrac{s_5-s_4}{s_5}\times100=\dfrac{45}{125}\times100=36\%$

If a particle is thrown with velocity more than

$10\ m/s$ vertically upward, then the distance traveled by the particle in last second of its ascent is:

0%
$g$
0%
$\dfrac { g }{ 2 }$
0%
$\dfrac { g }{ 4 }$
0%
$\dfrac { g }{ 8 }$

Explanation

time of ascent be t

$0=10-gt$

$t=1s$

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

A satellite orbiting round the earth appears stationary when:

0%
Its time period is one day and it is rotating in the same sense as that of earth.
0%
Its time period is one day and it is rotating normal to the direction of earth.
0%
Its time period is $12\ hr$ and it is rotating in the same direction as that of earth.
0%
Its time period is $12\ hr$ and it is rotating normal to earth.

A block of mass 5 kg is moving horizontally at a speed of 1.5 m/s. A perpendicular force of 5 N acts on it for 4 sec. What will be the shortest distance of the block from the point where the force started acting.

0%
10 m
0%
8 m
0%
6 m
0%
2 m

Explanation

Assume initial velocity of 1.5 m/s is in the x-direction

Since there are no forces on it in this direction, there will be no acceleration.

So, distance

$S_x=1.5\times 4=6m$

In the y-direction,

$F=5N$ and

$m=5kg$

Acceleration in y=direction,

$a_y=\dfrac{F}{m}=\dfrac{5}{5}=1 m/s^2$

$S_y=\dfrac{1}{2}a_yt^2=\dfrac{1}{2}\times 1\times 4^2=8m$

Resolving the x and y vector we get,

$S^2=S_x^2+S_y^2$

$S^2=6^2+8^2$

$S=\sqrt{36+64}$

$S=\sqrt{100}$

$S=10m$

A balloon is ascending vertically with an acceleration of 0.2 m

$s^{-2}$ . Two stones are dropper from it at an interval of 2 s. The distance between them when the second stone is dropped is (take g = 9.8 m

$s^{-2}$ ):

0%
24 m
0%
4.9 m
0%
19.6 m
0%
20.0 m

Explanation

A balloon is ascending vertically with an acceleration=

$0.2 m/s^2$

Two stones are dropped from it at an interval of

$t=2s$ .

$g=9.5 m/S$

The distance between them when the

$2nd$ stone is dropped:

$s_1=\cfrac {1}{2}at^2$

$(\because u=0)$

or,

$s_1=\cfrac {1}{2}\times 0.2\times 2^2$

$=\cfrac {1}{2}\times \cfrac {2}{10}\times 2^2$

$=\cfrac {4}{10}=0.4 m$ .

Due to free fall the

$1st$ stone will have,

$s=\cfrac {1}{2}\times (0.2+9.8)\times 2^2$

$=\cfrac {1}{2}\times 10 \times 4$

$=20$ .

$\therefore$ Hence the

$2nd$ stone is dropped=

$(20+4)=24 m$ .

The velocity acquired by a body moving with uniform acceleration is

$30 \,m/s$ in

$2$ seconds and

$60 \,m/s$ in

$4$ seconds. The initial velocity is:

0%
$Zero$
0%
$2 m/s$
0%
$4 m/s$
0%
$10 m/s$

Explanation

Let,

Constant acceleration, $a$

Initial velocity $u$

Apply kinematic Equation

$v=u+at$

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

Compare acceleration at both event at time, ${{t}_{1}}\,\And \,{{t}_{2}}$

$\dfrac{{{v}_{2}}-u}{{{t}_{1}}}=\dfrac{{{v}_{3}}-u}{{{t}_{2}}}$

$\Rightarrow \dfrac{30-u}{2}=\dfrac{60-u}{4}$

$\Rightarrow u=0$

Initial velocity $u=zero$

Two particles start moving from the same point along the same stright line. The first moves with constant velocity v and the second with constant acceleration a. During the time that elapses before the second catches the first, the greater distance between the particles is

0%
$\dfrac{v^2}{a}$
0%
$\dfrac{v^2}{2a}$
0%
$\dfrac{2v^2}{a}$
0%
$\dfrac{v^2}{3a}$

Explanation

A distance between the two particles is maximum when both have the same velocity

let the second move has a velocity v at time t

$v=at$

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

distance cover by the first paticle

$s_{1}$ is

$s_{1}=vt=v\times \dfrac{v}{a}=\dfrac{v^2}{a}$

distance cover the particle

$s_{2}$

$=\dfrac{1}{2}\times at^2=\dfrac{v^2}{2a}$

distance between both the particle is

$s_{1}-s_{2}$

$= \dfrac{v^2}{a}-\dfrac{v^2}{2a}=\dfrac{v^2}{2a}$

hence the B option is correct

A body lying initially at point (3, 7) starts moving with a constant acceleration of 4i. Its position after 3 s is given by the co-ordinates:

0%
(7, 3)
0%
(7, 18)
0%
(21, 7)
0%
(3, 7)

Explanation

The initial coordinate of the particle is

$(3,7)$

x-coordinate is 3 unit

y-coordinate is 7 unit

As the body starts from rest so initial velocity

$=0$

The acceleration of the body along x-axis

$=4 units$

The acceleration of the body along y-axis

$=0$

So, after 3 seconds,

Using equation of motion along x-axis,

$x=x_0+ut+\frac{1}{2}at^2$

$u=0\quad a=4units\quad t=3seconds$

$s=3+\frac{1}{2}\times4\times3^2=21units$

As there is no motion along y-axis so it will remain unchanged

Hence the final coordinate are

$(21,7)$

A body of mass

$500 g$ is thrown upward with a velocity

$20 m/s$ and reaches back to the surface of a planet after

$20$ sec Then the weight of the body on that planet is :

0%
$2N$
0%
$4 N$
0%
$5 N$
0%
$1 N$

Explanation

The acceleration when the mass is going up is given as,

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

${\left( {20} \right)^2} = {\left( 0 \right)^2} + 2as$

$a = \dfrac{{200}}{s}$ (1)

The acceleration when the mass is going down is given as,

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

$s = \dfrac{1}{2}a{\left( {10} \right)^2}$

$a = \dfrac{s}{{50}}$ (2)

From equation (1) and (2), we get

$\dfrac{{200}}{s} = \dfrac{s}{{50}}$

$s = 100\;{\rm{m}}$

Substitute the value of $s$ in the equation (2), we get

$a = \dfrac{{100}}{{50}}$

$= 2\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$

The weight is given as,

$W = 0.5 \times 2$

$= 1\;{\rm{N}}$

Thus, the weight of the body on that planet is $1\;{\rm{N}}$ .

If a car at rest accelerates uniformly to a speed of 144 km

$h^{-1}$ in 20 s, then it covers a distance of:

0%
20 m
0%
400 m
0%
1440 m
0%
2880 m

Explanation

Given

$u=0$

$v=144kmph=144\times\frac{5}{18}=40 m/s$

$t=20sec$

We know,

$v=u+at$

$40=a\times20$

$a=2m/s^2$

Now,

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

$s=\frac{1}{2}at^2=\frac{1}{2}\times2\times20^2=400m$

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

$10$ second then:

0%
$s_2=s_1$
0%
$s_2=2s_1$
0%
$s_2=3s_1$
0%
$s_2=4s_1$

Explanation

Distance traveled in first

$10 seconds$

$S_1 = u t + \dfrac{1}{2} a t^2 = 0 t + \dfrac{1}{2} a 10^2 = 50 a$

Total distance traveled in 20 seconds =

$S = u t + \dfrac{1}{2} a t^2 = 0 t + \dfrac{1}{2} a 2062 = 200 a$

Distance traveled from

$10s$ to

$20s$

$= S_2 = 200 a - 50 a = 150 a$

Hence answer is C.

$50-kg$ crate is being pushed across a horizontal floor by a horizontal force of

$575\ N$ . If the coefficient of sliding friction is

$0.25$ , what is the acceleration of the crate?

0%
Zero
0%
$1\ m/s^{2}$
0%
$3\ m/s^{2}$
0%
$6\ m/s^{2}$
0%
$9\ m/s^{2}$

Explanation

Given,

Force on crate, $F=\,575\,N$

Mass of crate, $m=50\,kg$

Coefficient of friction, $\mu =0.25$

From the equilibrium of forces

$ma=F-\mu mg$

$a=\dfrac{F-\mu mg}{m}$

$a=\dfrac{575-0.25\times 50\times 9.81}{50}$

$a=9.047\,\,m/{{s}^{2}}$

Acceleration of block is $9.047\,\,m/{{s}^{2}}$

978587_1065697_ans_0091dca6bfda42aca2249c5f8c091c33.jpg

An object moving with an speed of 25 m/s is decelerating at a rate given by

$a = -5\sqrt{v}$ , where v is speed at any time. The time taken by the object to come to rest is:

0%
1 s
0%
5 s
0%
2 s
0%
3 s

Explanation

We know that acceleration is time derivative of velocity so the given relation can be written as

$\dfrac{dv}{dt}=-5\sqrt[2]{v}$ or

$\dfrac{dv}{\sqrt[2]{v}}=-5dt$ integrating above equation we get

$2\sqrt[2]{v}=-5t+c$ .....(1)

where c is integrating constant.

To evaluate the constant just put

$t=0$ and

$v=25$ as given in the question.

By doing so we lead to

$c=10$ so the equation-1 becomes

$2\sqrt[2]{v}=-5t+10$ now put

$v=0$ in above equation we get

$t=2second$ .

Option C is correct.

A particle starts from rest, accelerates at

$2\ m/s^{2}$ for

$10s$ and then goes for constant speed for

$30s$ and then decelerates at

$4\ m/s^{2}$ till it stops. What is the distance travelled by it?

0%
$750\ m$
0%
$800\ m$
0%
$700\ m$
0%
$850\ m$

Explanation

A)
Initially particle starts from rest. So

$u=0 \ and \ a=2m/s^2 and t=10 sec$ and formula for this case is:

$s=ut+(1/2)a(t^2)$ and using this

$s = 0*10 + (1/2)*2*(10^2) = 100m$

and using the given information and this formula

v = u + at

we get the value of final velocity and on substituting all values we get:

0+2*10 = 20m/sec
Given that we have constant speed implies initial velocity is the same as the final velocity.

v=u and using this information and the formula

$v=u+at$

we get a=0 as t can’ t be zero. and

$t=30 sec$ (given). And from part

$u=20m/sec$

Now to get the value of distance in this part we use this formula :

$s=ut+(1/2)a(t^2)$ and using this we get:

$20*30+(1/2)*0*(30^2) = 600m .$

Now

$a=(-)4m/sec^2$

as particle is decelerating and given statement is it decelerates till is stops implies that final velocity is zero. And from above part

$u=20m/sec$ and to get the value of distance we use this formula:

$v^2-u^2 = 2as$ and using this we get:

$0-20^2 = 2*(-4)*s => s =400/8 = 50m.$

So using all the part info we get total distance travelled: 100+600+50 = 750

Two particle P and Q are initially

$40m$ apart P behind Q . Particle P starts moving with a uniform velocity

$10m/s$ towards Q . Particle Q starting from rest has an acceleration

$2ms^2$ in the direction of velocity of P. Then the minimum distance between P and Q will be

0%
$45m$
0%
$15m$
0%
$35m$
0%
$30m$

Two bikes

$A$ and

$B$ start from a point. A move uniform speed

$40\ m/s$

$B$ starts from rest with uniform acceleration

$2\ m/s^{2}$ . If

$B$ starts at

$t = 0$ and

$A$ starts from the same point at

$t = 10\ s$ , then the time interval during the journey in which

$A$ was ahead of

$B$ is

0%
$20\ s$
0%
$8\ s$
0%
$10\ s$
0%
$A$ is never ahead of $B$

A man throws balls with the same speed vertically upwards one after the other at an interval of

$2$ second. What should be the speed of the throw so that more than two balls are in sky at any time? (Given

$g=9.8$

${m/s^2}$ )

0%
More than $19.6$ ${m/s}$
0%
At least $9.8$ ${m/s}$
0%
Any speed less than $19.6$ ${m/s}$
0%
Only with speed $19.6$ ${m/s}$

A mass of

$3\ kg$ descending vertically downward support a mass of

$2\ kg$ by means of a light string passing over a pulley. At the end of

$5\ s$ the string breaks. How much high from now the

$2\ kg$ mass will go ? (

$g=9.8\ m/s^{2}$ )

0%
$4.9\ m$
0%
$9.8\ m$
0%
$16.9\ m$
0%
$2.45\ m$

Explanation

Given that,

Mass ${{m}_{1}}=3\,kg$

Mass ${{m}_{2}}=2\,kg$

Now, the initial acceleration a of the system

$a=\dfrac{3g-2g}{3+2}$

$a=\dfrac{g}{5}$

At $5\ sec$

Now, from equation of motion

$v=u+at$

$v=0+\dfrac{g}{5}\times 5$

$v=9.8\,m/s$

When the string breaks at $t= 5\ s$ ,

The mass $2\ kg$ more under gravity

So, height is

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

$h=\dfrac{{{\left( 9.8 \right)}^{2}}}{2\times 9.8}$

$h=4.9\,m$

Hence, the height is $4.9\ m$

1032302_1080866_ans_38a174b65d3741b89ac4c33b903519a7.PNG

A force

$\vec {F}=(2\hat {i}+\hat {j}+3\hat {k})\ N$ acts on a particle of mass

$1\ kg$ for

$2s$ . If initial velocity of particle is

$u=(2\hat {j}+\hat {i}m/s$ . Speed of particle at the end of

$2s$ will be

0%
$12\ m/s$
0%
$6\ m/s$
0%
$9\ m/s$
0%
$4\ m/s$

Explanation

Given,

Initial velocity, $u=\left( 2\hat{j}+\hat{k} \right)\,m/s$

Force, $F=\left( 2\hat{i}+\hat{j}+3\hat{k} \right)\,N$

$\vec{a}=\dfrac{{\vec{F}}}{m}=\dfrac{2\hat{i}+\hat{j}+3\hat{k}}{1}=\left( 2\hat{i}+\hat{j}+3\hat{k} \right)\,m/{{s}^{2}}$

$\vec{v}=\vec{u}+\vec{a}t$

$\vec{v}=\left( 2\hat{j}+\hat{k} \right)+\left( 2\hat{i}+\hat{j}+3\hat{k} \right)\times 2$

$\vec{v}=\left( 4\hat{i}+4\hat{j}+7\hat{k} \right)\,m/s$

$\left| {\vec{v}} \right|=\sqrt{{{4}^{2}}+{{4}^{2}}+{{7}^{2}}}=9\,m/s$

Final velocity is $9\,m/s$

A particle moves along a straight line with a uniform acceleration

$-5m/s^{2}$ and initial velocity

$12.25/s$ distance traveled by it in

$3rd$ second

0%
0.6725m
0%
zero
0%
0.25m
0%
0.75m

Explanation

Distance travelled in

$n^{th}$ second is given by:-

$x_n=(4+an)-\dfrac {a}{2}$

for

$n=3, u=12.25\ m/s,\ a=-5\ m/s^2$

$x_3=(12.25-3\times 5)+\dfrac {5}{2}=0.25\ m$

A stone is thrown vertically upwards. On its way, passes a point P with a speed

$v$ and point Q is

$20 m$ higher than Q with a speed

$\cfrac{\sqrt{5}}{3}v$ . The value of speed: [

$g = 10 m/s^2$ ]

0%
$20\sqrt5 m/s$
0%
$15\sqrt5 m/s$
0%
$25 m/s$
0%
$30 m/s$

Explanation

The velocity at P is V

Heigth difference is 20m

Velocity at Q

$=\dfrac{\sqrt4 v}{3}$

$(\dfrac{\sqrt5 v}{3})^2=v^2-2\times 20\times 10$

$2\times 20\times 10=v^2-\dfrac{5}{9}v^2$

$\dfrac{4}{9}v^2=2\times 20\times 10$

$\dfrac{2}{3}v=2\times 10$

$v=\dfrac{2\times 10\times 3}{2}=30 m/s$

A ball is thrown upwards with a speed of

$50 m/s$ . Find the distance travelled by the ball in last

$2 s$ of its ascent. (Take

$10 m/s^2$ )

0%
$35 m$
0%
$10 m$
0%
$20 m$
0%
$5 m$

Explanation

Initial velocity

$-50 m/s$ (opposite to the direction of acceleration)

Final velocity

$=10 m/s$ (completing the ascent)

Time taken

$=\dfrac{0-(-50)}{acceleration}=\dfrac{50}{10}=5s$

$S_{th}=$ Initial acceleration +

$\dfrac{1}{2}$ acceleration (

$2\times$ (time)-1)

$S_{4th}=-50+\dfrac{1}{2}\times 10(7)=-15 m$

$S_{5th}=-50+\dfrac{1}{2}\times 10(9)=-5 m$

$S_{4th}+S_{5th}=-20 m$ (Negative as it is opposite to direction of acceleration)

A parachutist after bailing out falls 80 m without air resistance. Then the parachute is opened, and he decelerates at a uniform rate of

$2 m/s^2$ . If he reaches the ground with a speed of

$4 m/s$ , then the height at which he bailed out is [

$g = 10 m/s^2$ ]

0%
$576 m$
0%
$676 m$
0%
$476 m$
0%
$376 m$

Explanation

Final velocity after falling 80m,

$V_s^2=0+2\times 10\times 80(v_c=0)$

$V_s=\sqrt{1600}$

$V_{s1}=40 m/s$

Final velocity when reaching ground =4 m/s

$(V_{s2})^2=V_{s1}^2-2ah$

$4^2=40^2-4h$

$h=\dfrac{(40-4)(40+4)}{4}$

$h=9\times 44=396$

Total height of bailing

$=396+80=476 m$

CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 8 - 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 8 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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