A particle moves along a straight line such that its displacement S varies with time t as S = a + bt + g$$t^2$$ Column I Column II i Acceleration at t = 2 s a $$\beta + 5 \gamma$$ ii Average velocity during third second b $$2\gamma$$ iii Velocity at t = 1 s c $$\alpha$$ iv Initial displacement d $$\beta = 2 \gamma$$

(i) - b, (ii) - a, (iii) - d, (iv) - c.

(i) - b, (ii) - a, (iii) - d, (iv) - d.

(i) - a, (ii) - b, (iii) - d, (iv) - c.

(i) - b, (ii) - c, (iii) - d, (iv) - a.

MCQ Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 12

The ratio of time taken by two cars P and Q starting from rest moving along a straight road with equal accelerations is

$\sqrt{2} : 1$ , then the :

0%
final velocity of car P > final velocity or car Q.
0%
final velocity of car P < final velocity of car Q.
0%
ratio of $V_P$ to $V_Q$ is 2 : $\sqrt{2}$
0%
ratio of distance travelled by car 'P' to car 'Q' is 2 :1

Explanation

The ratio of time

${t_P}: {t_Q}= \sqrt2 :1$ , acceleration is same, Let acceleration is

$a$ , Initial velocity

$u=0$ .

Using

$v=u+at$ ,

$V_P=0+at_p$ ....1

$V_Q= 0+at_Q$ .....2

Subtracting equations 2 from 1,

$V_P-V_Q= a(t_P-t_Q)$ =

$a(\sqrt2-1)$ >0 so Option(A) is correct.

Divide equations 1 by 2,

$\dfrac{V_P}{V_Q}= \dfrac{\sqrt2}{1}$

$\Rightarrow V_P : V_Q= 2 :\sqrt2$ so option (C) is correct.

Using

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

$S_P= \dfrac{1}{2}at_p^2$ ......3

$S_Q=\dfrac{1}{2}at_Q^2$ .....4

Divide equation 3 by 4,

$\Rightarrow \dfrac{S_P}{S_Q}= \dfrac{t_P^2}{t_Q^2}$

$\Rightarrow S_P : S_Q= 2:1$ So option (D) is correct.

At a certain acceleration, a racing car takes

$5 s$ to double its speed while moving at

$2 m{s}^{-1}$ . If now, it moves at

$4 m{s}^{-1}$ , at the same acceleration, what time will it take to double its speed?

0%
$5s$
0%
$10s$
0%
$15s$
0%
None of these

Explanation

Let the acceleration be

$a$ .

$v=u+at$

$4=2+(a\times 5)$

$a=\dfrac{2}{5}ms^{-2}$
Let the time required be

$T$ .

$\therefore 8=4+(\dfrac{2}{5}\times T)$

$4=\dfrac{2}{5}\times T$

$T=10\ s$

A person drops a metallic sphere from the top of a tower of height 125 m. Another person at a distance of 80 m from the foot of the tower hears the sound of the sphere hitting the ground after a time interval of 5.25 s. Find the velocity of the sound in air.( $g = 10 {ms}^{-2}$ )

0%
$140{ms}^{-1}$
0%
$280{ms}^{-1}$
0%
$320{ms}^{-1}$
0%
$460{ms}^{-1}$

A block is thrown with a velocity of

$2 m/s$ (relative to ground) on a belt, which is moving with velocity

$4 m/s$ in opposite direction of the initial velocity of block. If the block stops slipping on the belt after

$4 s$ of the throwing then choose the correct statement(s)

0%
Displacement with respect to ground is zero after $2.66$ and displacement with respect to ground is $12 m$ after $4 s$
0%
Displacement with respect to ground in $4 s$ is $4 m$ .
0%
Displacement with respect to belt in $4 s$ is $-12 m$ .
0%
Displacement with respect to ground is zero in $\dfrac{8}{3} s$ .

Explanation

Let belt is moving towards left (4m/s) and initial velocity of block is towards right(2m/s).

Now let us assume everthing wrt belt velocity of block will be 6m/s towards right and final velocity will be zero.

So acceleration is

$1.5m/s^2$ because time taken to stop wrt belt is 4 sec.

Distance travelled wrt belt is

$S=6\times4-\frac{1}{2}\times1.5\times4^2=12$

Distance travelled by belt in that time is

$4\times4=16$

Now taking direction of movement of belt as positive we can say that

Option C. Displacement wrt to belt is -12m.

Option B. Displacement wrt ground is (-12+16=4)=4m.

Also displacement wrt ground will be zero when

$S=6\times t-\frac{1}{2}\times1.5\times t^2-4\times t=0$

$=>t=\frac{8}{3}$ so option D is also correct.

The car drives straight off the edge of a cliff that is

$57 m$ high. The investigator at the scene of the accident notes that the point of impact is

$130 m$ from the base of the cliff. How fast was the car traveling when it went over the cliff? (in m/s)

0%
$36$
0%
$37$
0%
$38$
0%
$39$

Explanation

To get the time of flight, used the fact that as the car falls over the cliff the initial vertical component of its velocity is zero.

It then falls under gravity with an acceleration of g;

$\therefore S=\cfrac { 1 }{ 2 } g{ t }^{ 2 }$

$\therefore { t }^{ 2 }=\cfrac { 2s }{ g }$

$t=\sqrt { \cfrac { 2s }{ g } } =\sqrt { \cfrac { 2\times 57 }{ 9.8 } } =3.41s$

Now we know the time of flight, so horizontal component of velocity as it went over the cliff;-

$v=\cfrac { \triangle s }{ \triangle t } =\cfrac { 130 }{ 3.41 } =38㎧$

A particle is projected upwards with some velocity. At what height from ground should another particle be just dropped at the same time so that both reach the ground simultaneously. Assume that first particle reaches to a maximum height

$H$ .

0%
$3H$
0%
$4H$
0%
$5H$
0%
$6H$

Explanation

Let the height from which the second particle is dropped be

$h$ .
Time of flight of second particle

$T = \sqrt{\dfrac{2h}{g}}$
Since, both particles hit the ground simultaneously. So both must have equal time of flight.
Time of flight of first particle

$T = \dfrac{2u\sin\theta}{g}$
Using equation for maximum height

$H = \dfrac{u^2\sin^2\theta}{2g}$
We get

$H = \dfrac{g}{8}\times T^2 = \dfrac{g}{8}\times \dfrac{2h}{g}$

$\implies \ h = 4H$

How long does the race take? (in s)

0%
$199.8$
0%
$180$
0%
$125.4$
0%
$190.8$

Explanation

If L is the length of the race then the tortoise run this race with time:

${ t }_{ tortoise }=\cfrac { L }{ { v }_{ tortoise } } =\cfrac { L }{ 0.1 } \rightarrow 1$

When the tortoise crosses the finish line the hare is

$20㎝=0.2m$ behind tortoise. It means that during time

${ t }_{ tortoise }$ it traveled the distance

$L-0.2$

Since he stopped for rest of

$3$ minutes

$=3\times 603=180s$

and his speed is

$10{ v }_{ tortoise }=1㎧$

$\therefore L-0.2=10{ v }_{ tortoise }\left( { t }_{ tortoise }-180 \right)$

Putting

$L=0.1{ t }_{ tortoise }$ from

$1$

$0.1{ t }_{ tortoise }-0.2=1\left( { t }_{ tortoise }-180 \right)$

${ t }_{ tortoise }=199.8s$

You are driving along the street at the speed limit (

$35mph$ ) and

$50$ meters before reaching a traffic light you notice it becoming yellow. You accelerate to make the traffic light within the

$3$ seconds it takes for it to turn red. What is your speed as you cross the intersection? Assume that the acceleration is constant and that there is no air resistance.

0%
$30 mph$
0%
$40 mph$
0%
$50 mph$
0%
$60 mph$

Explanation

This is the motion with constant acceleration. If the acceleration of the car is

$a$ then we can write the expression for traveled distance.

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

Here,

$u=$ initial velocity

$=35mph$

$=15.645㎧$

After

$3$ sec, the car travels distance

$50$ meters. Then we can find acceleration

$a=\cfrac { 2(s-{ v }_{ 0 }t) }{ { t }^{ 2 } }$

$=\cfrac { 2\left( 50-15.645\times 3 \right) }{ { 3 }^{ 2 } }$

$=\cfrac { 2(50-46.935) }{ 9 }$

$=0.6811㎨$

From relation,

$v=u+at$

$=15.6+0.68\times 3$

$=17.64㎧$

i.e.,

$=40mph$

Who wins the race?

0%
Tortoise
0%
Hare
0%
Both reach simultaneously
0%
Cannot be judged

Explanation

Traveled time of tortoise

$t=\cfrac { s }{ v } =\cfrac { 1000 }{ 0.2 } =5000s$

Traveled time of rabbit;-

$1$ . He runs the first

$200m$ at

$2㎧$

$\therefore { t }_{ 1 }=\cfrac { s }{ v } =\cfrac { 200 }{ 2 } =100s$

$2$ . Then he take a nap for

$1.3h$

${ t }_{ 2 }=1.3h=1.3\times 3600s$

$=4680s$

$3$ . Then he run the last

$800m$ with speed

$3㎧$

${ t }_{ 3 }=\cfrac { s }{ v } =\cfrac { 800 }{ 3 } =267s$

Total traveled time of rabbit

$={ t }_{ 1 }+{ t }_{ 2 }+{ t }_{ 3 }$

$=100+4680+267$

$=5047s$

Since time taken by rabbit is more than time taken by tortoise. Tortoise wins the race by

$47s$ .

Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of

$444 m/s$ in

$1.83$ seconds, then what is the distance that the sled travels? (in m)

0%
$406 m$
0%
$40 m$
0%
$46 m$
0%
$506 m$

Explanation

Given, initial velocity

$u=0 m/s,$ final velocity

$v=444 m/s$ and time

$t=1.83 s$

$a$ be the acceleration.

Using

$v=u+at$ ,

we get

$444=0+a(1.83)$

$\implies$

$a=444/1.83=242.62\sim 243 m/s^2$

$d$ be the traveled distance by sled.

Using formula

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

$444^2-0^2=2(243)d$

$\implies$

$d=405.63 \sim 406 m$

upward velocity the rock was shot at (in m/s)

0%
$19.6$
0%
$20$
0%
$21$
0%
$23$

Explanation

Here free fall motion of the rock. The equations, which describe the motion of the rock are the following;-

$y(t)={ y }_{ 0 }=+{ v }_{ i }t-9.8\cfrac { { t }^{ 2 } }{ 2 }$

$v(t)={ v }_{ i }-9.8t$

${ v }^{ 2 }\left( t \right) -{ v }_{ i }^{ 2 }=-2\times 9.8\left[ y(t)-{ y }_{ 0 } \right]$

where,

${ v }_{ i }=$ initial velocity

${ y }_{ 0 }=$ initial height (height of the building)

At the maximum height the velocity of the rock is zero,

Then from the second equation we can fine the initial velocity (since

$t=2s$ )

${ v }_{ i }=9.8t$

$=9.8\times 2=19.6㎧$

A body is thrown vertically up from the ground with a speed

$v_{initial}$ and it reaches maximum height h at time

$t=t_o$ . What is the height to which it would have risen at time

$t=t_o/2$

0%
$0.75h$
0%
$h$
0%
$0.5h$
0%
$0.3h$

Explanation

In this problem we have free fall motion

$y(t)={ y }_{ initial }+{ v }_{ initial }t-4.9{ t }^{ 2 }$

$v(t)={ v }_{ initial }-9.8t$

${ v }^{ 2 }\left( t \right) -{ v }_{ initial }^{ 2 }=-19.6\left[ y(t)-{ y }_{ initial } \right]$

The initial height of the object is

$0$ (ground level)

$\therefore y(t)={ v }_{ initial }t-4.9{ t }^{ 2 }\rightarrow 1$

$v(t)={ v }_{ initial }-9.8t\rightarrow 2$

${ v }^{ 2 }\left( t \right) -{ v }_{ initial }^{ 2 }=-19.6(t)\rightarrow 3$

Case I: The object reaches the maximum height h at

${ t }_{ 0 }$ . At the maximum height the velocity of the object is zero and

$y\left( { t }_{ 0 } \right) =h$

$h={ v }_{ initial }{ t }_{ 0 }-4.9{ t }_{ 0 }^{ 2 }\rightarrow 4$

$0={ v }_{ initial }-9.8{ t }_{ 0 }\rightarrow 5$

${ v }_{ initial }^{ 2 }=19.6h$

Case II: Height of the object at

$t=\cfrac { { t }_{ 0 } }{ 2 }$

Substitute this time in equation

$1$ and

$3$

$y\left( { { t }_{ 0 } }/{ 2 } \right) =\cfrac { { v }_{ initial }{ t }_{ 0 } }{ 2 } =4.9\cfrac { { t }_{ 0 }^{ 2 } }{ 4 } \rightarrow 7$

$v\left( { { t }_{ 0 } }/{ 2 } \right) ={ v }_{ initial }-\cfrac { 9.8{ t }_{ 0 } }{ 2 } \rightarrow 8$

${ v }^{ 2 }\left( { { t }_{ 0 } }/{ 2 } \right) -{ v }_{ initial }^{ 2 }=-19.6y\left( { { t }_{ 0 } }/{ 2 } \right) \rightarrow 9$

We know,

${ v }_{ initial }={ t }_{ 0 }9.8$

Substitute in equation

$4$ and

$7$

$h=9.8{ t }_{ 0 }^{ 2 }-4.9{ t }_{ 0 }^{ 2 }=4.9{ t }_{ 0 }^{ 2 }\rightarrow 10$

$y\left( { { t }_{ 0 } }/{ 2 } \right) =\cfrac { 9.8{ t }_{ 0 }^{ 2 } }{ 4 } =4.9\cfrac { { t }_{ 0 }^{ 2 } }{ 4 }$

$=3.675{ t }_{ 0 }^{ 2 }\rightarrow 11$

From equation

$10$

${ t }_{ 0 }^{ 2 }=\cfrac { h }{ 4.9 }$

Substitute this in equation

$11$

$y\left( { { t }_{ 0 } }/{ 2 } \right) =3.675{ t }_{ 0 }^{ 2 }$

$=3.675\times \cfrac { h }{ 4.9 }$

$=0.75h$

From the top of a tower of height

$40m$ , a ball is projected upwards with a speed of

$20m/s$ at an angle of elevation of

$30^0$ . The ratio of the total time taken by the ball to hit the ground to its time of flight (time taken to come back to the same elevation) is :

0%
$2:1$
0%
$1:2$
0%
$1:1$
0%
$3:2$

Explanation

Time of flight,

$T=\cfrac { 2v\sin { \theta } }{ g } =\cfrac { 2\times 20\times \sin { 30° } }{ 10 } =2sec$
Total time taken by the ball to hit the ground is

$t=\cfrac { distance }{ velocity } =\cfrac { 40 }{ 20\sin { 30° } } =\cfrac { 40 }{ 20\times \cfrac { 1 }{ 2 } } =4sec$
So, the ratio is

$4:2=2:1$

A car and a bike start racing in a straight line. The distance of the finish line from the starting line is

$100m$ . The minimum acceleration of the car for it to win, if it accelerates uniformly starting from rest and the bike moves with a constant velocity of

$10m/s$ , is

0%
$0.5{m}/s^{2}$
0%
$2{m}/s^{2}$
0%
$1{m}/s^{2}$
0%
$3{m}/s^{2}$

Explanation

For bike :

Distance cover to finish the race

$d = 100 \ m$

Velocity

$v = 10 \ m/s$

Time taken

$t = \dfrac{d}{v} = \dfrac{100}{10} = 10 \ s$

For car :

Let the acceleration be

$a$ .

Initial velocity of car

$u = 0$

Using

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

where

$t = 10 \ s$

So,

$100 = 0+\dfrac{1}{2}\times a\times 10^2$

$\implies \ a = 2 \ m/s^2$

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 particle of mass

$2\ m$ is projected at an angle

$45^{\circ}$ with the horizontal with a velocity of

$20\sqrt {2}m/ s$ . After

$1s$ , explosion takes place and the particle is broken into the two equal pieces. As a result of explosion, one part comes to rest. The maximum height from the ground attained by the other part is

0%
$50\ m$
0%
$25\ m$
0%
$40\ m$
0%
$35\ m$

Explanation

Given: Initial velocity

$u_{0} = 20\sqrt {2}m/s$ ; angle of projection

$\theta = 45^{\circ}$
Therefore horizontal and vertical components of initial velocity are

$u_{x} = 20\sqrt {2}\cos 45^{\circ} = 20\ m/s$
and

$u_{y} = 20\sqrt {2}\sin 45^{\circ} = 20\ m/s$
After

$1s$ , horizontal component remains unchanged while the vertical component becomes

$v_{y} = u_{y} - gt$
Due to explosion, one part comes to rest.
Hence, from the conservation of linear momentum, vertical component of second part will become

$v_{y}' = 20 m/s$ .
Therefore, maximum height attained by the second part will be

$H = h_{1} + h_{2}$
Using,

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

$\Rightarrow h_{1} = (20\times 1) - \dfrac {1}{2} \times 10\times (1)^{2} = 15\ m$

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

$h_{2} = \dfrac {v'^{2}_{y}}{2g} = \dfrac {(20)^{2}}{2\times 10} = 20 m$

$H = 20 + 15 = 35\ m$ .

A particle starts moving along a line from rest and comes to rest after moving distance

$d$ . During its motion, it had a constant acceleration

$f$ over

$2/3$ of the distance and covered the rest of the distance with constant retardation. The time taken to cover the distance is:

0%
$\sqrt {\dfrac{2d}{3f}}$
0%
$2\sqrt {\dfrac{d}{3f}}$
0%
$\sqrt {\dfrac{3d}{f}}$
0%
$\sqrt {\dfrac{3d}{2f}}$

Explanation

Let velocity at B be

$v$ .

From A to B :

Initial velocity,

$u = 0$

Acceleration,

$a= f$

Time taken,

$t=t_1$

Using Newton's first equation of motion,

$v=u+at$

Velocity at B,

$v_1 = 0+ft_1 = ft_1$

$v_1= ft_1$ ....(1)

Using Newton's second equation of motion,

$s=ut+\dfrac 12 at^2$

Distance covered,

$s= \dfrac{2d}{3}$

$\dfrac{2d}{3} = 0+\dfrac{1}{2}ft_1^2$

$\implies \ t_1 = \sqrt{\dfrac{4d}{3f}}$

$\implies \ t_1 =2 \sqrt{\dfrac{d}{3f}}$ .....(b)

From B to C :

Initial velocity,

$u = v_1$

Final velocity,

$v = 0$

Acceleration,

$a$

Time taken,

$t=t_2$

Using Newton's first equation of motion,

$v=u+at$

$0 = v_1+at_2$

$\implies = v = -at_2$ ....(2)

From (1) and (2), we get

$a = -\dfrac{ft_1}{t_2}$

Using Newton's third equation of motion

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

$2a\dfrac{d}{3}=0^2-v_1^2$

$\dfrac{d}{3} = \dfrac{-v^2}{2a} = \dfrac{-(ft_1)^2}{2\times (-ft_1/t_2)} = \dfrac{ft_1t_2}{2}$ ....(a)

From (a) and (b),

$\dfrac{d}{3} = \dfrac{ft_2}{2}\times \sqrt{\dfrac{4d}{3f}}$

$\implies \ t_2 = \sqrt{\dfrac{d}{3f}}$

Thus total time taken,

$t = t_1+t_2 = 2 \sqrt{\dfrac{d}{3f}}+\sqrt{\dfrac{d}{3f}} = 3\sqrt { \dfrac{d}{3f}}$

$t=\sqrt { \dfrac{3d}{f}}$

2085264_738276_ans_bbdc7c25dd544fbc9c699fbef528e70f.png

On a horizontal flat ground, a person is standing at a point

$A$ . At this point, he installs a

$5 m$ long pole vertically. Now, he moves

$5 m$ towards east and then

$2 m$ towards pole to the top of the second pole. Find the displacement and magnitude of the displacement of the bird.

0%
$5\hat{i}+2\hat{j}-2\hat{k}$ ; $\sqrt{33}m$
0%
$6\hat{i}+2\hat{j}-2\hat{k}$ ; $\sqrt{23}m$
0%
$5\hat{i}+3\hat{j}-2\hat{k}$ ; $\sqrt{33}m$
0%
$5\hat{i}+2\hat{j}-2\hat{k}$ ; $\sqrt{23}m$

A train of length / = 350 m starts moving rectilinearly with constant acceleration a =

$3.0 \times 10^{-2}$

$ms^{-2}$ . After r = 30 s from start, the locomotive headlight is switched on (event 1), and 60 s after this event, the tail signal light is switched on (event 2).
a.Find the distance between these events in the reference frame fixed to the train and to the Earth.
b.How and at what constant velocity v relative to the Earth must a certain reference frame R move for the two events to occur in it at the same point?

0%
a. 350 m is with respect to train; 242 m is with respect to Earth; b. 4.03 m/s
0%
a. 350 m is with respect to train; 252 m is with respect to Earth; b. 4.03 m/s
0%
a. 350 m is with respect to train; 262 m is with respect to Earth; b. 4.03 m/s
0%
a. 250 m is with respect to train; 242 m is with respect to Earth; b. 4.03 m/s

A car accelerates from rest at a constant rate

$\alpha$ for some time after which it decelerates at a constant rate

$\beta$ and comes to rest. It total time elapsed is

$t$ , then maximum velocity acquired by car will be:

0%
$\cfrac { \left( { \alpha }^{ 2 }-{ \beta }^{ 2 } \right) t }{ \alpha \beta }$
0%
$\cfrac { \left( { \alpha }^{ 2 }+{ \beta }^{ 2 } \right) t }{ \alpha \beta }$
0%
$\cfrac { \left( { \alpha }^{ }+{ \beta }^{ } \right) t }{ \alpha \beta }$
0%
$\cfrac { \left( { \alpha }^{ }{ \beta }^{ } \right) t }{ \alpha +\beta }$

Explanation

$\textbf{Step 1 - Calculation of acceleration and deacceleration :}$

Apply first equation of motion,

From rest, car accelerates for time

$t$ ,

$V=u+at=0+\alpha t$

$\alpha = \dfrac {V}{t_{1}}$

$....(1)$

Let maximum velocity is

$V$ .

The car begins to deaccelerate after obtaining maximum velocity

$0=V-\beta (t-t_{1})$

$\therefore \beta = \dfrac {V}{t - t_{1}}$

$....(2)$

$\textbf{Step 2 - Calculation of velocity :}$

By adding equation

$(1)$ and

$(2)$

$t = V\left (\dfrac {1}{\alpha} + \dfrac {1}{\beta}\right )$

$V = \dfrac {\alpha \beta t}{\alpha + \beta}$

Hence, the maximum velocity acquired by car is

$V = \dfrac {\alpha \beta t}{\alpha + \beta}$

Two cars are travelling towards each other on a straight road at velocities

$15ms^{-1}$ and

$16ms^{-1}$ respectively. When they are

$150$ m apart, both the drivers apply the breaks and the cars decelerates at

$3ms^{-2}$ and

$4ms^{-2}$ until they stop. How for apart will they be when they have come to a stop.

0%
$86.5$ m
0%
$89.5$ m
0%
$85.5$ m
0%
$81.36$ m

Explanation

solving with respect to the car

$1$ moving with speed

$15m/s$ .

using concept of relative motion we get

$V_{21}=V_{2g}-V_{1g}$

same with acceleration

$a_{21}=a_{2g}-a_{1g}$

so velocity car

$2$ is

$15+16=31m/s, acc=4+3=7m/s^2$ and

$separation =150m$

distance covered before coming to rest

$\frac{31^2}{2\times7}=S=>S=68.64$

so now separation between them is

$150-68.64=81.36m$

so none of the option is correct.

A man running with a uniform speed 'u' on a straight road observes a stationary bus at a distance 'd' ahead of him. At that instance , the bus starts with an acceleration 'a' .The condition that he would be able to catch the bus is:

0%
$\displaystyle d \, \leq \, \frac{u^2}{a}$
0%
$\displaystyle d \, \leq \, \frac{u^2}{2a}$
0%
$\displaystyle d \, \leq \, \frac{u^2}{3a}$
0%
$\displaystyle d \, \leq \, \frac{u^2}{4a}$

Explanation

Relative speed of man w.r.t bus

$=u$

and acceleration w.r.t bus

$=-a$

if man catch the bus

$d\leq x$ where x is calculated as

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

$v=$ final velocity w.r.t bus that is zero.

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

$x=\frac{u^{2}}{2a}$

hence

$d \leq \dfrac{u^{2}}{2a}$

1007220_878117_ans_cae1f62e8eae4b61aa1698f21e42e5a1.png

A body is falling freely from a point

$A$ at a certain height from the ground and passes through points

$B,C$ and

$D$ (vertically as shown) so that

$BC=CD$ . The time taken by the particle to move from

$B$ to

$C$ is

$2$ seconds and from

$C$ to

$D$

$1$ second. Time taken to move from

$A$ to

$B$ in seconds is

0%
$0.6$
0%
$0.5$
0%
$0.2$
0%
$0.4$

Explanation

Velocity of the body at point A

$u_A = 0$
Let the velocity of body at points B and C be

$u_B$ and

$u_C$ respectively.
Let

$BC = CD = S$
For journey

$A$ to

$B$ :
Using

$u_B =u_A +gt_{AB}$
where

$t_{AB}$ is the time taken to move from A to B.

$\therefore$

$u_B = 10t_{AB}$ .......(1)

$(\because u_A = 0)$
For journey

$B$ to

$C$ :
Using

$u_C = u_B +gt_{BC}$

$\therefore$

$u_C = u_B+20$
Using

$S = u_Bt_{BC}+\dfrac{10}{2}t_{BC}^2$
where

$t_{BC} = 2$
We get

$S = 2u_B+5(2)^2 = 2u_A+20$ .....(2)
For journey

$C$ to

$D$ :
Using

$S = u_Ct_{CD}+\dfrac{10}{2}t_{CD}^2$
where

$t_{CD} = 1$
We get

$S = u_C+5(1)^2 = u_C+5$ .....(3)
Equation (2) - (1), we get

$0 = 2u_B-u_C + 15$

$\implies \ 2u_B = u_C-15$
Or

$2u_B = u_B+20-15$

$\implies \ u_B = 5$
From (1), we get

$5 = 10t_{AB}$

$\implies \ t_{AB} = 0.5 \ s$

The maximum height reached by the ball as measured from the ground would be:

0%
$52\ m$
0%
$31.25\ m$
0%
$83.25\ m$
0%
$82.56\ m$

Explanation

Initial velocity,

$u=15+10=25$ m/s.

at maximum height,

$V=0$

$\therefore$

$V=u+at$

$0=25-10t$

$\Rightarrow$

$t=2.5s$

Time to reach on the floor

$t_0=2.13 \ sec$

Since

$t<t_0,$ , therefore as per the question ball is at

its maximum height at 2.13s.

We have,

$x={ x }_{ 0 }+ut+\dfrac { 1 }{ 2 } { at }^{ 2 }$

Taking ground as reference,

${ x }_{ 0 }=50+2=52$

$\therefore$

${ x }_{ max }=52+25\times 2.5+\dfrac { 1 }{ 2 } \times \left( -10 \right) \times { 2.13 }^{ 2 }$

$=82.56m$

t0=ug=2510=2.5s

The maximum separation between the floor of elevator and the ball during its flight is:

0%
$30\ m$
0%
$15\ m$
0%
$7.62\ m$
0%
$9.5\ m$

Explanation

Initial point of floor is considered as reference point.

We have,

$x={ x }_{ 0 }+ut+\dfrac { 1 }{ 2 } { at }^{ 2 }$

${ x }_{ b }=2+25t-5{ t }^{ 2 }$

${ x }_{ ele }=10t+{ 1/2\times 10t }^{ 2 }=10t+{5t }^{ 2 }$

Separation,

$\Delta ={ x }_{ b }-{ x }_{ ele }$

$\Rightarrow \Delta =2+15t-10{ t }^{ 2 }$

$\dfrac { d\Delta }{ dt } =0\Rightarrow 15-20t=0$

$\Rightarrow t=\dfrac { 3 }{ 4 }$

$\dfrac { { d }^{ 2 }\Delta }{ { dt }^{ 2 } } <0$ ,

$\Delta$ at

$t=\dfrac { 3 }{ 4 }$ will be max.

$\therefore$

$\Delta$ at

$t=\dfrac { 3 }{ 4 } =2+15\times \dfrac { 3 }{ 4 } -10\times { \left( \dfrac { 3 }{ 4 } \right) }^{ 2 }$

$=7.625m$

At the initial moment, three points A, B, and C are on a horizontal straight line (east-west line) with equal distances between adjacent points. Point A begins to move northward with a constant velocity u and point C towards south at a constant acceleration a without any initial velocity. How should point B move towards north so that all the three points always remain on one straight line? The points begin to move simultaneously.

0%
Point B moves with the initial velocity 2u directed towards north and a constant acceleration a/2 directed south.
0%
Point B moves with the initial velocity u/2 directed towards north and a constant acceleration a/2 directed south.
0%
Point B moves with the initial velocity u/3 directed towards north and a constant acceleration a/2 directed south.
0%
Point B moves with the initial velocity u directed towards north and a constant acceleration a/2 directed south.

The distance travelled by the ball during its flight is:

0%
$32.64\ m$
0%
$31.86\ m$
0%
$52\ m$
0%
$30.56\ m$

Explanation

Let initial point of floor of elevator be the reference point.

We have,

$x={ x }_{ 0 }+ut+\dfrac { 1 }{ 2 } { at }^{ 2 }$

$\therefore$

${ x }_{ ball }=2+25t-\dfrac { 1 }{ 2 } \times 10\times { t }^{ 2 }\quad \rightarrow \left( 1 \right)$

${ x }_{ elevator }=0+10t+5{ t }^{ 2 }\quad \rightarrow \left( 2 \right)$

The ball will hit the floor when

${ x }_{ ball }={ x }_{ elevator }$

On solving

$(1)$ &

$(2)$ ,

$t=2.13s$

$\therefore$ Distance traveled by the ball during height is,

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

$=25\times 2.13-\dfrac { 1 }{ 2 } \times 10\times { 2.13 }^{ 2 }$

$=30.56m$

A particle of

$m = 5 kg$ is momentarily at rest at

$x = 0$ at

$t = 0$ . It is acted upon by two forces

$\vec{F}_{1}$ and

$\vec{F}_{2}$ .

$\vec{F}_{1} = 70 \hat{j} N$ . The direction and magnitude of

$\vec{F}_{2}$ are unknown. The particle experiences a constant acceleration, a, in the direction as shown in figure. Neglect gravity.
a. Find the missing force

$\vec{F}_{2}$ .
b. What is the velocity vector of the particle at

$t=10s$ ?
c. What third force,

$\vec{F}_{3}$ , is required to make the acceleration of the particle zero? Either give magnitude and direction of

$\vec{F}_{3}$ or its components.

0%
a. $30\sqrt{2}$
b. $60\hat{i} + 80\hat{j}$
c. $30\hat{i} - 40\hat{j}$
0%
a. $30\sqrt{2}$
b. $60\hat{i} + 90\hat{j}$
c. $30\hat{i} - 40\hat{j}$
0%
a. $30\sqrt{2}$
b. $60\hat{i} + 80\hat{j}$
c. $30\hat{i} - 50\hat{j}$
0%
a. $30\sqrt{2}$
b. $60\hat{i} + 80\hat{j}$
c. $40\hat{i} - 40\hat{j}$

A ship is sailing due north at a speed of

$1.25$ m s

$^{-1}$ . The current is taking it towards east at the rate of

$2$ m s

$^{-1}$ and a sailor is climbing a vertical pole in the ship at the rate of

$0.25$ m s

$^{-1}$ . Find the magnitude of the velocity of the sailor with respect to ground.

0%
$=\dfrac{3}{2}\sqrt{\dfrac{5}{2}}ms^{-1}$
0%
$=\dfrac{3}{22}\sqrt{\dfrac{5}{2}}ms^{-1}$
0%
$=\dfrac{23}{2}\sqrt{\dfrac{5}{2}}ms^{-1}$
0%
$=\dfrac{13}{2}\sqrt{\dfrac{5}{2}}ms^{-1}$

$B_1$ ,

$B_2$ , and

$B_3$ are three balloons ascending with velocities

$v, 2v$ , and

$3v$ , respectively. If a bomb is dropped from each when they are at the same height, then

0%
Bomb from $B_1$ reaches the ground first
0%
Bomb from $B_2$ reaches the ground first
0%
Bomb from $B_3$ reaches the ground first
0%
They reach the ground simultaneously

A particle moves along a straight line such that its displacement S varies with time t as S = a + bt + g

$t^2$

	Column I		Column II
i	Acceleration at t = 2 s	a	$\beta + 5 \gamma$
ii	Average velocity during third second	b	$2\gamma$
iii	Velocity at t = 1 s	c	$\alpha$
iv	Initial displacement	d	$\beta = 2 \gamma$

0%
(i) - b, (ii) - a, (iii) - d, (iv) - d.
0%
(i) - b, (ii) - a, (iii) - d, (iv) - c.
0%
(i) - a, (ii) - b, (iii) - d, (iv) - c.
0%
(i) - b, (ii) - c, (iii) - d, (iv) - a.

A flat car of mass

$M$ is at rest on a frictionless floor with a child of mass

$m$ standing at its edge. If the child jumps off from the car toward right with an initial velocity

$u$ , with respect to the car, velocity of the car after its jump.

0%
$\nu = \dfrac{mu}{2m + M}$
0%
$\nu = \dfrac{mu}{m + M}$
0%
$\nu = \dfrac{2mu}{3m + M}$
0%
$\nu = \dfrac{mu}{m + 4M}$

In a car race, car A takes 4 s less than car B at the finish and passes the finishing point with a velocity v more than the car B. Assuming that the cars start from rest and travel with constant accelerations

$a_1 = 4 ms^{-2}$ and

$a_2 = 1 ms^{-2}$ respectively, find the velocity of v in m/s.

0%
9
0%
8
0%
10
0%
11

A cat, on seeing a rat at a distance d = 5 m, starts with velocity u = 5 m/s and moves with the acceleration

$\alpha = 2.5 ms^{-2}$ in order to catch it, while the rate with the acceleration

$\beta$ starts from rest. For what value of

$\beta$ will the cat overtake the rat? (in

$ms^{-2}$ )

0%
5
0%
6
0%
7
0%
8

A person is travelling from the ground floor to the first floor in a mall using an escalator. Consider the following three separate cases. When the person is standing on the moving escalator it takes one minute for him to reach the first floor. If the escalator does not move it takes him

$3$ minutes to walk up the stationary escalator to reach the top. How long will it take for the person to reach the top if he walks up the moving escalator?

0%
$30\ \sec$
0%
$45\ \sec$
0%
$40\ \sec$
0%
$35\ \sec$

A ball is thrown downwards with a speed of 20 m/s from the top of a building 150 m high and simultaneously another ball is thrown vertically upwards with a speed of 30 m/s from the foot of the building. Find the time after which both the balls will meet. (g = 10

$ms^{-2}$ )

0%
33
0%
3
0%
4
0%
5

An ideal spring having force constant k is suspended from the rigid support and a block of mass M is attached to its lower end. The mass is released from the natural length of the spring. Then the maximum extension in the spring is

0%
4Mg/k
0%
2Mg/k
0%
Mg/k
0%
Mg/2k

The physical situation in List I with graphs of the variation of total energy (

$E$ ), potential energy (

$U$ ) and kinetic energy (

$K$ ) with time in List II are given. Match the statements from List I with those in List II and select the correct answer using the code given below the lists.

List I	List II
(P) A mass attached to an unstretched ideal spring, released from position O in vertical plane from rest until it reaches its maximum extension. Assume that the gravitational P.E reference is at the equilibrium position of $m$
(Q) An object of mass is released from tower AB undergoes free fall
(R) An object of mass $m$ is being pulled by a constant force in horizontal direction on a horizontal frictionless surface
(S) A particle of mass $m$ is given velocity ${v}_{0}$ downward on a rough inclined surface whose coefficient of friction is $\mu=\tan{\theta}$

0%
P- 1; Q- 4; R- 2; S- 3
0%
P- 2; Q- 3; R- 1; S- 4
0%
P- 4; Q- 1; R- 2; S- 3
0%
P- 4; Q- 1; R- 3; S- 2

The pulley and string are shown in Fig. smooth and of negligible mass. For the system to remain in equilibrium, the angle

$\theta$ should be

0%
$0^0$
0%
$30^0$
0%
$45^0$
0%
$60^0$

Explanation

We know that,

Tension on the string

$=mg$

$\longrightarrow \left( i \right)$

The component of

$T$ which is working

$\Rightarrow T\cos\theta$

also,

$T\cos\theta =\sqrt { 2 } mg$

Since there are

$2$ blocks,

$2T\cos\theta =\sqrt { 2 } mg$

$2\left( mg \right) \cos\theta =\sqrt { 2 } mg$ (from (i))

$\cos\theta =\dfrac { 1 }{ \sqrt { 2 } }$

$\theta ={ \cos }^{ -1 }\dfrac { 1 }{ \sqrt { 2 } }$

[C]

$\boxed { \theta ={ 45 }^{ 0 } }$

Let

$t$ be a time instant greater than

${t}_{1}$ but less than

$3\ s$ . The velocity at this time instant is :

0%
$4\ (t-{t}_{1})$
0%
$2\ {t}_{1}+4\ (t-{t}_{1})$
0%
$2\ {t}_{1}-4\ (t-{t}_{1})$
0%
$4t$

A traveller while in a uniformly moving train throws a ball up in the air. The ball will return-

0%
In his hand
0%
Ahead in the direction of motion of the train
0%
Trail behind
0%
Deflected sideways

During a rainy day, rain is falling vertically with a velocity

$2\ ms^{-1}$ . A boy at rest starts his motion with a constant acceleration of

$2ms^{-2}$ along a straight road. Then the rate at which the angle of the axis of umbrella with the vertical should be changed so that the rain always appears to fall parallel to the axis of umbrella is

0%
$\dfrac{4}{4+t^{2}}$
0%
$\dfrac{3}{3+t^{2}}$
0%
$\dfrac{2}{2+t^{2}}$
0%
$\dfrac{1}{1+t^{2}}$

A force of

$25\ N$ acts on a body at rest for

$0.2s$ and a force of

$70\ N$ acts for the next

$0.1s$ in opposite direction. If the final velocity of the body is

$5ms^{-1}$ , the mass of the body is:

0%
$1\ kg$
0%
$2\ kg$
0%
$0.8\ kg$
0%
$0.4\ kg$

Explanation

$25=ma_1$

$u=a_1t=\dfrac{25}{m}0.2$

$70=ma_2$

$v=u-a_t$

$=\dfrac{25}{m}0.2$ -

$\dfrac{70}{m}0.1$

$m=0.4kg$

A stone falls from rest. The total distance covered by it in the last second of its motion is equal to the distance covered in the first three seconds. What is the height from which the stone was dropped? Take g = 10

$m/s^2$ :

0%
45 m
0%
100 m
0%
125 m
0%
200 m

Explanation

Distance covered in 3 sec

$=\dfrac{1}{2}gt^2=\dfrac{1}{2}\times10\times 3^2=45m$

At the end of 'n' second its velocity

$u=gn$

At the end of 'n+1' second its velocity

$v=g(n+1)$

Distance traveled in the (n+1)th second

$=\dfrac{(2n+1)g}{2}=\dfrac{(20n+10)}{2}=10n+5$

$10n+5=45\Rightarrow (n+1)=5$

Distance covered in 5 second

$=\dfrac{1}{2}gt^2=\dfrac{1}{2}\times10\times 5^2=125m$

Two blocks

$(A)\ 2\ kg$ and

$(A)\ 5\ kg$ rest one over the other on a smooth horizontal plane. The coefficient of static and dynamic friction between

$(A)$ and

$(B)$ is the same and equal to

$0.60$ . The maximum horizontal force

$F$ that can be applied to

$(B)$ in order that both

$(A)$ and

$(B)$ do not have any relative motion is :

0%
$42\ N$
0%
$84\ kgf$
0%
$5.4\ kgf$
0%
$1.2\ N$

A motorist travelling at

$54\ km/hr$ when he observes a traffic signal at a distance of

$280\ m$ ahead turning red at time

$t=0$ . If he decides to decelerate & then accelerates by the same magnitude, so as to pass the signal at

$54\ km/hr$ at

$t=28$ seconds, when the signal turns green, then

0%
Magnitude of deceleration is $\dfrac{5}{7}m/s^{2}$
0%
Magnitude of deceleration is $1 m/s^{2}$
0%
Minimum speed reached is $5\ m/s$
0%
Minimum speed reached is $10\ m/s$

Explanation

$u=54km/hr=54\times \dfrac{1000}{3600}m/s$

$u=15m/s$

Now, from

$A$ to

$B$

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

$140=15\times 14-\dfrac{1}{2}a\times (14)^{2}$

$a=\dfrac{70}{98}m/s^{2}=\dfrac{5}{7}m/s^{2}$

$V_{B}=u-at$

$V=15-\dfrac{5}{7}\times 14=10m/s$ .

Two towns are connected through a straight highway. A car runs between two towns with constant speed

$36\ km/hr$ . One day track of

$100m$ is to be covered with speed

$18\ km/hr$ . For this car accelerates and retards at rate

$1\ { m/sec }^{ 2 }$ . By what time the car will get delayed that day. (When car reaches at any town its speed is

$36\ km/hr$ .)

0%
$8.5\ s$
0%
$10.5\ s$
0%
$12.5\ s$
0%
$6.5\ s$

Particle has initial velocity

$9\ ms^{-1}$ due east and constant acceleration of

$2\ ms^{-2}$ due west. If the distance covered by it in fifth second of its motion is

$\dfrac{n}{10}m$ , then the value of

$'n'$ is:

0%
$5$
0%
$1$
0%
$3$
0%
$2$

Explanation

Here the particle velocity will become zero in 4.5 seconds,as

$v=u+at$

$0=9-2t$

$t=4.5sec$

Displacement in the same period will be

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

$-81=2\times (-2)s$

$s_{4.5}=20.25m$

Let us call the point of velocity being 0 as Point A

Now body will start moving backwards,

Displacement in next 0.5 sec from point A,

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

$s_{0.5}=\frac{1}{2}\times(-2)\times 0.25=-0.25m$

Hence distance covered is 0.25m

(negative sign indicates body moved in backward direction. It is to be noted that displacement and distance are not to be confused to be same)

Distance traveled in 4 seconds

$s_4=36-16=20m$

Distance covered in 5th second is

$D=s_{4.5}+s_{0.5}-s_4$

$D=20.25+0.25-20$

$D=0.5m$

which means the distance covered by particle in 5th second of its motion is

$\frac{n}{10}=\frac{5}{10}=0.5$

Therefore value of n is 5

For a particle undergoing rectilinear motion with uniform acceleration , the magnitude of displacement is one third the distance covered in some time interval. The magnitude of final velocity is less than magnitude if initial velocity for this time interval. Then the ratio of initial speed to the final speed for this time interval is:

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

A stone thrown upwards with speed

$u$ attains maximum height

$h$ . Another stone thrown upwards from the same point with sapped

$2u$ attains maximum height

$H$ . What is the relation between

$h$ and

$H$ ?

0%
$2h=H$
0%
$3h=H$
0%
$4h=H$
0%
$5h=H$

Explanation

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

$H=\dfrac{(2u)^2}{2g}$

$H=4h$

CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 12 - 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 12 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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