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

It moves with uniform acceleration

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

A particle moves half the time of its journey with velocity u. The rest of the half time it moves with two velocities

$v_1$ and

$v_2$ such that half the distance it covers with

$v_1$ and the other half with

$v_2$ . Find the net average velocity assume straight line motion:

0%
$\dfrac { u\left( v_{ 1 }+v_{ 2 } \right) +2v_{ 1 }v_{ 2 } }{ 2\left( v_{ 1 }+v_{ 2 } \right) }$
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$\dfrac { 2u\left( v_{ 1 }+v_{ 2 } \right) }{ 2u+v_{ 1 }+v_{ 2 } }$
0%
$\dfrac { u\left( v_{ 1 }+v_{ 2 } \right) }{ 2v_{ 1 } }$
0%
$\dfrac { 2v_{ 1 }v_{ 2 } }{ u+v_{ 1 }+v_{ 2 } }$

Explanation

$S_{ 1 }=\dfrac { ut }{ 2 } ,\dfrac { S_{ 2 } }{ 2 } =v_{ 1 }t_{ 1 }=v_{ 2 }t_{ 2 }$ (i)

$\dfrac{t}{2}=t_1+t_2$ (ii)
From (i) and (ii)

$\Rightarrow t_1=\dfrac{v_2t}{2(v_1+v_2)}$

$v_{av}=\dfrac{S}{t}=\dfrac{S_1+S_2}{t}=\dfrac { \dfrac { ut }{ 2 } +{ 2 }v_{ 1 }t_{ 1 } }{ t }$
Put the value of

$t_1$ and get

$v_{av}=\dfrac { u\left( v_{ 1 }+v_{ 2 } \right) +2v_{ 1 }v_{ 2 } }{ 2\left( v_{ 1 }+v_{ 2 } \right) }$
1573805_1734797_ans_d716f4c2b00849f0b53c843d8fe8960d.png

The velocity of a particle varies with time as shown below. The distance travelled by the particle during

$t=2s$ and

$t=6s$ is:

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$2\pi m$
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$(2\pi +40 m)$
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$4\pi m$
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$40 m$

The maximum velocity attained by the car is

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$\dfrac { \alpha \beta }{ 2\left( \alpha +\beta \right) } t$
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$\dfrac { \alpha \beta }{ \alpha +\beta } t$
0%
$\dfrac { 2\alpha \beta }{ \alpha +\beta } t$
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$\dfrac { 4\alpha \beta }{ \alpha +\beta } t$

Explanation

From A to B, applying

$v = u + at$ , we get

$v_0=0+at\Rightarrow t_1=v_0/a$
From B to C, again applying v = u + at, we get

$0=v_0-bt_2\Rightarrow t_2=v_0/b$
Given

$t_1+t_2=t\Rightarrow$

$\dfrac{v_0}{\alpha }+\dfrac{v_0}{\beta }$

$=t\Rightarrow v_0=\dfrac{\alpha \beta t}{\alpha+\beta}$

$v_0$ is the maximum velocity attained.
1572680_1734738_ans_53d899017a784092be91949d81d3a043.png

The time after which they are closet to each other

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1/3 s
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8/3 s
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1/5 s
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8/5 s

The velocity of the body at any instant is

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$\dfrac { M+{ 2 }Nt^{ 4 } }{ 4 }$
0%
$2N$
0%
$\dfrac { M+{ 2 }N }{ 4 }$
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$2Nt^3$

Explanation

$4s=m+2nt^{4}$

$s=\dfrac{m+2nt^{4}}{4}$

$s=\dfrac{m}{4}+\dfrac{2nt^{4}}{4}$

$v=\dfrac{ds}{dt}=\dfrac{dm}{dt^{4}}+\dfrac{2nt^{4}}{4dt}$

$\dfrac{ds}{dt}=0+\dfrac{4.2nt^{3}}{4}$ Differentiation of any constant is 0

$v=\dfrac{ds}{dt}=2nt^{3}$

A particle is found to be at rest when seen from a

$S_1$ and moving with a constant velocity when seen from another frame

$S_2$ select the possible options:

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Both the frames are non-inertial
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$S_1$ is inertial and $S_2$ is non-inertial
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Both the frames are inertial
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$S_1$ is non-inertial and $S_2$ is inertial

The distance covered by the second body when they meet is

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8 m
0%
16 m
0%
24 m
0%
32 m

Explanation

Let at time t, both bodies will meet.

distance traveled by body A = distance traveled by body B

$u_1* t = u_2t + \dfrac{1}{2} at^2$

$8* t = 0 * t + \dfrac{1}{2}* 4 *t^2$

$8t = 2 t^2$

$t = 4 s$

Distance covered by body B

$= u_2t^2 + \dfrac{1}{2}at^2 \\ = 0*t + \dfrac{1}{2} * 4 *(4)^2 \\ = 32 m$

Option 'D' is correct.

The velocity of the body at the end of 1 s from the start is

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$2N$
0%
$\dfrac{M+2N}{4}$
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$2(M+N)$
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$\dfrac{2M+N}{4}$

Explanation

$s=\dfrac{m+2Nt^4}{4}$

$\Rightarrow v=\dfrac{ds}{dt}=2Nt^3$
Putting

$t=1 s$ , we get

$v = 2N$

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

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uniform acceleration
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uniform speed
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variable speed
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variable acceleration

Explanation

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

Thus, speed

$=$ constant

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

1628542_1773478_ans_8fc22fdb5d9643f9acadc512e9d5add2.png

A stone is dropped from the top of the tower . Its speed after it has fallen

$20\,m$ is

(Take

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

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$-10\, ms^{-1}$
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$10\, ms^{-1}$
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$-20\, ms^{-1}$
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$20\, ms^{-1}$

Explanation

Here ,

${\text{u}}$ = Initial velocity =

$0$

${\text{v}}$ = Final velocity

${\text{s}}$ = Distance travelled =

$20\,m$

${\text{a}}$ = acceleration = g =

$10 \, ms^{-2}$

Now , we know that from the equation of motion that ,

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

${\text{v}} = \sqrt{{\text{u}}^2 + 2as}$

${\text{v}} = \sqrt{ 0 + 2 \times 10 \times 20}$

${\text{v}} = \sqrt{400} \, ms^{-1}$

${\text{v}} = 20 \, ms^{-1}$

Starting from rest at the top of an inclined plane a body reaches the bottom of the inclined plane in 4 second. In what time dies the body cover one-fourth the distance starting from rest at the top?

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

Explanation

Let

$s$ be the total distance and

$a$ be the acceleration

Now, we know that initially, the body is at rest
Now, from the equations of motion, we get

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

$s=\cfrac { 1 }{ 2 } a\times {4}^{2}....(2)$

To find the time taken for covering

$\cfrac{s}{4}$ , we put

$s=\cfrac{s}{4}$ in equation (1)

So,

$\cfrac{s}{4}=\cfrac{1}{2}a\times {t}^{2}....(3)$

Now dividing equation

$2$ by

$3$ , we get

$t=2s$

An iron sphere of mass

$10\ kg$ has the same diameter as an aluminium sphere of mass is

$3.5\ kg$ . Both spheres are dropped simultaneously from a tower. When they are

$10\ m$ above the ground, the have the same

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acceleration
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momenta
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potential energy
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kinetic energy

Explanation

$(a)$ acceleration
Explanation: Momentum, potential energy and kinetic energy depend on the mass of the object; as well as on some other factors. But acceleration, in this case, is the acceleration due to gravity; which does not depend on mass or velocity. So, option (a) is correct.

A book lying on a table is an example of

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a body at rest
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a body in motion
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a body neither at rest nor motion
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none of these

A body falls from rest in the gravitational field of the earth. The distance travelled in the fifth second of its motion is

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

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$25 \,m$
0%
$45\,m$
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$90\,m$
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$125\,m$

Explanation

Since the initial velocity is

$0$ ,the distance in

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

$h_n = \dfrac{g}{2} (2n - 1) \, \Rightarrow h_{5^{th}} = \dfrac{10}{2} (2 \times 5 - 1) = 45 m$

Rewrite the sentence using proper alternative.
The total energy of an falling freely toward the ground ____

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decreases
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remains unchanged
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increases
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increases in the beginning and then decreases

On another planet, a marble is released from rest at the top of a high cliff. It falls

$4.00 m$ in the first

$1 s$ of its motion. Through what additional distance does it fall in the next

$1 s$ ?

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$4.00m$
0%
$8.00m$
0%
$12.0m$
0%
$16.0m$
0%
$20.0m$

Explanation

We take downward as the positive direction with

$y = 0$ and

$t = 0$ at the top of the cliff. The freely falling marble then has

$v_{0} = 0$ and its displacement at

$t = 1.00 s$ is

$Δy = 4.00 m$ . To find its acceleration, we use

$y=y_{0}+v_{0}t+\frac{1}{2}at^{2}\rightarrow (y-y_{0})=\Delta y=\frac{1}{2}at^{2}\rightarrow a=\frac{2\Delta y}{t^{2}}$

$a=\frac{2(4.00m)}{(1.00s)^{2}}=8.00m/s^{2}$

The displacement of the marble (from its initial position) at

$t = 2.00 s$ is found from

$\Delta y=\frac{1}{2}at^{2}$

$\Delta y=\frac{1}{2}(8.00m/s^{2})(2.00s)^{2}=16.0m$

The distance the marble has fallen in the

$1.00 s$ interval from

$t = 1.00 s$ to

$t = 2.00 s$ is then

$∆y = 16.0 m − 4.0 m = 12.0 m$ .

and the answer is (c).

An electron with a speed of

$3.00 \times 10^6 m/s$ moves into
a uniform electric field of magnitude

$1.00 \times10^3 \,N/C$ . The field lines are parallel to the electrons velocity and pointing in the same direction as the velocity. How far does the electron travel before it is brought to rest?

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$2.56 \,cm$
0%
$5.12 \,cm$
0%
$11.2 \,cm$
0%
$3.34 \,m$
0%
$4.24 \,m$

Explanation

The electric force is opposite to the field direction, so it is opposite to the velocity of the electron. From Newton's second law, the acceleration the electron will be

$a_x=\dfrac{F_x}{m}=\dfrac{qE_x}{m}=\dfrac{(-1.60\times 10^{-19}C)(1.00\times 10^3 \,N/C}{9.11\times10^{-31} Kg}$

$-1.76\times 10^{14} \,m/s^2$
The kinematics equation

$v_x^2=v_{0x}^{2}+2a_x(\Delta x)$ , with

$v_x=0$ , gives the stopping distance as

$\Delta x=\dfrac{-v_{0x}^2}{2a_x}=\dfrac{-(3.00\times 10^6 \,m/s^2)}{2(1.76\times 10^{14} \,m/s^2)}=2.56\times 10^{-2} \,m=2.56m$

Relative to another coordinate system

$\mathrm{S}'$ (denoted by single prime) moving with a vertical velocity

$\vec{\mathrm{v}}_{0}$ , the equation of motion of the object becomes

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$\displaystyle \mathrm{m}(\frac{\mathrm{d}\vec{\mathrm{v}}}{\mathrm{d}\mathrm{t}})=-\mathrm{m}\vec{\mathrm{g}}-\mathrm{b}\vec{\mathrm{v}}$
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$m\left ( \frac{d\vec{\mathrm{v}}}{dt} \right )=-m\vec{g}-b(\vec{\mathrm{v}}-\vec{\mathrm{v_{0}}})$
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$m\left [ \left ( \frac{d\vec{\mathrm{v}}}{dt}-\vec{\mathrm{v_{0}}} \right ) \right ]=-m\vec{g}-b(\vec{\mathrm{v}}+\vec{\mathrm{v_{0}}})$
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$m\left ( \frac{d\vec{\mathrm{v}}}{dt} \right )=m(\vec{g}-\vec{\mathrm{v_{0}}})-b\vec{\mathrm{v}}$

Explanation

The acceleration due to gravity will be the same in both systems

$\vec{v}+\vec{v_{0}}=\vec{v'}$

$\Rightarrow \vec{v}=\vec{v'}-\vec{v_{0}}$

$\frac{d\vec{v}}{dt}=\frac{d\vec{v}}{dt}-\frac{d\vec{v_{0}}}{dt}$

$\frac{d\vec{v}}{dt}=\frac{d\vec{v'}}{dt}$

$\Rightarrow m\frac{d\vec{v}}{dt}=m(\frac{d\vec{v'}}{dt})$

$\Rightarrow m\left ( \frac{d\vec{v'}}{dt} \right )=-mg-b(\vec{v'-\vec{v_{0}}})$

A particle moves with a non zero initial velocity

${ v }_{ 0 }$ and retardation

$kv$ , where

$v$ is the velocity at any time

$t$ .

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The particle will cover a total distance $\cfrac {{ v }_{ 0 }}{k}$
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The particle comes to rest at $t=\cfrac{1}{k}$
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Particle continues to move for long time
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at time $\cfrac{1}{\alpha}, v=\cfrac {{ v }_{ 0 }}{2}$

Explanation

$u=v_0,$ retardation

$a=kv$ and

$v$ is the velocity at time

$t$ .
using

$v=u-at,$ we will get

$v=v_0-kvt \Rightarrow t=\dfrac{v_0}{kv}-\frac{1}{k}$
if

$v=0, t =\infty$ . So the particle will be moved for a long time.

A body of mass m is dropped from a height of h. Simultaneously another body of mass 2m is thrown up vertically with such a velocity v that they collide at height

$\dfrac {h}{2}$ . If the collision is perfectly inelastic, the velocity of combined mass at the time of collision with the ground will be-

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$\sqrt {\dfrac {5gh}{4}}$
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$\sqrt {gh}$
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$\sqrt {\dfrac {gh}{4}}$
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None of these

Explanation

m mass falls a distance

$\dfrac {h}{2}$ in time, say t. Then,

$\dfrac {h}{2}=\frac {1}{2}gt^2\therefore t=\sqrt {\dfrac {h}{g}}$
For mass 2m,

$\dfrac {h}{2}=vt-\dfrac {1}{2}gt^2=v\sqrt {\dfrac {h}{g}}-\dfrac {h}{2}$

$\therefore v=\sqrt {gh}$
Now, let u is the velocity with which combined mass collides with ground, then

$u=\sqrt {(\dfrac {2}{3}v)^2+2g(\dfrac {h}{3})}$

$=\sqrt {\dfrac {4}{9}gh+\dfrac {2}{3} gh}$

$=\dfrac {\sqrt {10}}{3}\sqrt {gh}$

Since the collision takes place in vertical direction and external force is not zero in this direction, momentum conservation cannot be applied.
If the gravitational force( weight) is much less than the impulse force, it may be ignored and conservation of momentum can be applied.

A particle is projected with a velocity

$u$ in horizontal direction as shown in the figure. Find

$u$ (approx.) so that the particle collides orthogonally with the inclined plane of the fixed wedge.

0%
$10 \ m/s$
0%
$20 \ m/s$
0%
$10\sqrt{2}\ m/s$
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None of these

Explanation

Considering plane along Wedge axis,
when particle collides with the plane, its component of velocity parallel to inclined plane should be zero.

$\Rightarrow u \cos\theta - g \sin\theta \times t =0$ {Equation along the plane}

$t=\displaystyle \dfrac { u }{ g \tan \theta }$

$u=\displaystyle gt\times \tan\theta = \dfrac{gt}{\sqrt{3}}$
Now,
Consider the plane along original x-y axis,
Thus,
Horizontal displacement before it hits the plane

$=ut$
vertical displacement during the time t

$=\dfrac{1}{2} g t^2$
Now ,

$\tan 30^\circ{} =\displaystyle \dfrac{ h- \frac{1}{2} g t^2 } { ut }$

$\dfrac{1}{\sqrt{3}} = \displaystyle \dfrac{ h- \dfrac{1}{2} g t^2 } { ut }$
Hence ,

$\dfrac{1}{3} = \displaystyle ( \dfrac{h} {gt^2} - \dfrac{1}{2} )$

$t=\displaystyle \sqrt{ \dfrac{6h}{5g}}$
Hence ,

$t = 2.5938$ s
So,

$u = gt \, \tan 30^\circ{} = \displaystyle \dfrac{9.8 \times 2.5938}{\sqrt 3} = 14.69 \simeq 10 \sqrt{2}$ m/s

The displacement of a particle moving in a straight line is given by

$x=16t-2t^{2}$ (where,

$x$ is in meters and

$t$ is in second). The distance traveled by the particle in

$8$ seconds [starting from

$t$

$=$ 0] is

0%
$24 m$
0%
$40 m$
0%
$64 m$
0%
$80 m$

Explanation

Given:

$x=16 t - 2t^{2}$

Comparing with second equation of motion:

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

It can be concluded that:

$u=16\ m/s,\quad a=4\ m/s^2$

Now, using first equation of motion:

$v = u+at$

$v=16 - 4t$
For

$v = 0$ we get,

$t= 4 s$

Here, direction of velocity will reverse. So total distance travelled will be the sum of displacement in first four seconds and displacement in next four seconds.

Now,
Displacement at

$t=0,\quad x=0$

$t=4,\quad x=32$

$t=8,\quad x=0$

$\therefore$ Distance

$=$ (distance from

$x=0$ to

$x=32$ ) + (distance from

$x=32$ to

$x=0$ )

$\therefore$ Distance

$= 32+32 =64\ m$

A water tap leaks such that water drops fall at regular intervals. Tap is fixed

$5\ m$ above the ground. First drop reaches the ground and at that very instant third drop leaves the tap. At this instant the second drop is at a height of

0%
$3\ m$
0%
$4.5\ m$
0%
$3.75\ m$
0%
$2.5\ m$

Explanation

Velocity of drop 5m below the tap can be obtained by energy conservation.
Assume that potential energy is zero 5m below the tap.
Now,

$v=\sqrt { 2gh } =10\quad m/s$
Now, time taken to achieve this velocity

$=\dfrac { 10 }{ 10 } =1s$
it is given that drops leave the tap at a regular time interval. Thus, second drop will leave at 0.5s as the third drop is leaving at 1s.
Now, total time of motion of second drop is 0.5s
Thus, distance moved by second drop in 0.5s is

$s=\dfrac { 1 }{ 2 } \times 10\times { (0.5) }^{ 2 }=1.25$
Thus height from ground level is

$5-1.25=3.75m$

A ball is thrown vertically upwards (relative to the train) in a compartment of a moving train. The face of the person sitting inside the compartment is towards engine of the train.

0%
The ball will maintain the same horizontal velocity as that of the person (or the compartment) at the time of throwing.
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If the train is accelerating then the horizontal velocity of the ball will be different from that of the train velocity, at the time of throwing.
0%
If the ball appears to be moving backward to the person sitting in the compartment it means that speed of the train is increasing.
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If the ball appears to be moving ahead of the person sitting in the compartment it means the train's motion is retarding.

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

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

What can be concluded about the motion of the body?

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

A circus girl throws three rings upwards one after the other at equal intervals of half a second. She catches the first ring half second after the third was thrown. Then,
(

$\mathrm{g}=$ acceleration due to gravity)

0%
the velocity of projection of rings is $\displaystyle \frac{3\mathrm{g}}{4}$
0%
the maximum height attained by rings is $\displaystyle \frac{\mathrm{g}}{32}$
0%
when the first ring returns to her hand, the second ring was coming downwards and it is on the height of $\displaystyle \frac{\mathrm{g}}{4}$ (from the ground).
0%
when the first ring returns to her hand, the third ring was going up and has travelled a distance of $\displaystyle \frac{\mathrm{g}}{4}$ (from the ground).

Explanation

The circus girl catches the first ring after

$\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}$ second

$\Rightarrow$ the ring was in air for

$\dfrac{3}{2}$ second in which for first

$\dfrac{3}{4}$ second it was going up and for the next

$\dfrac{3}{4}$ second it was coming down.
Let the height attained by first ring be h in

$\dfrac{3}{4}$ second, obviously, its velocity at the instant will be zero.
if u be the velocity of projection of rings, then

$0=u-g\left ( \dfrac{3}{4} \right )\Rightarrow u=\dfrac{3g}{4}$

$\Rightarrow (A)$ is correct.
The maximum height h attained by the ring is given by,

$0=\left ( \dfrac{3g}{4} \right )^{2}-2gh$

$\Rightarrow h=\dfrac{9g}{32}\Rightarrow (B)$ is not correct.
When the first ring returns to her hand, the second ring has traveled for 1 second. Of that 1 second,

$\dfrac{3}{4}$ seconds will be spent in going up and for the rest it is coming down.

$\Rightarrow h'=\dfrac{1}{2}g\left ( \dfrac{1}{4} \right )^{2}=\dfrac{g}{32}$

$\Rightarrow$ Distance travelled by second ring by that instant is

$=\dfrac{9g}{32}-\dfrac{g}{32}=\dfrac{g}{4}$

$\Rightarrow$ (C) is correct.
The third ring has travelled for

$t=\dfrac{1}{2}$ second by that instant. it will be going up.
Corresponding distance from ground

$t=\left ( \dfrac{3g}{4} \right )\left ( \dfrac{1}{2} \right )-\dfrac{1}{2}g\left ( \dfrac{1}{2} \right )^{2}$

$=\dfrac{3g}{8}-\dfrac{g}{8}=\dfrac{g}{4}$

$\Rightarrow (D)$ is correct.

Two identical balls are shot upward one after another at an interval of

$2\ s$ along the same vertical line with same initial velocity of

$40\ ms^{-1}$ The height at which the balls collide is

0%
$50\ m$
0%
$75\ m$
0%
$100\ m$
0%
$125\ m$

Explanation

time taken for ball to rest is

$v= u +at$

$0 = 40 -gt$

$t = 4 s$

therefore ball 1 is at rest at height

$S = 40 \times 4 -\dfrac{1}{2}\times 10 \times 4^2 = 80 m$

velocity of ball 2 after 2s as it was thrown after the delay of 2s

$v = u+at$

$v = 40 - 10 \times 2 = 20 m/s$

$S_2 = 40 \times 2 - \dfrac{1}{2} \times 10 \times 2^2 = 60 m$

now ball 1 will drop with initial velocity of 0 and acceleration g and ball 2 with initial velocity of 20 m/s and deceleration of g and the separation is 20m

$S_1 = 0 + \dfrac{1}{2}gt^2$

$S_2 = 20t - \dfrac{1}{2}g t^2$

$S_1+S_2 = 20$

$\dfrac{1}{2}gt^2 + 20t - \dfrac{1}{2}gt^2 = 20$

$t = 1s$

$S_1 = 5m$

$S_2 = 15m$

hence ball 2 will be at 75m when they collide.

A particle is thrown upwards from ground. It experiences a constant resistance force which can produce a retardation of

$\displaystyle 2\ ms^{-2}.$ The ratio of time of ascent to time of descent is (g=

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

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

Explanation

$v = u + at$

$0 = u + -(g + 2)t$

$t = \dfrac{u}{g + 2}$

$0 - u^2 = 2as$

$- u^2 = 2 \times -(g + 2)s$

$s = u^2 / 2(g + 2)$

$s = \dfrac{1}{2} aT^2$

$\dfrac{u^2}{2(g+2)} = \dfrac{1}{2} \times (g-2)T^2$

$\dfrac{u}{\sqrt(g^2 - 4)} = T$

$\dfrac{t}{T}=\dfrac{\sqrt{g-2}}{\sqrt{g+2}}=\sqrt{\dfrac{8}{12}}=\sqrt{\dfrac{2}{3}}$

A person sitting in the rear end of the compartment throws a ball towards the front end.The ball follows a parabolic path.The train is moving with velocity of

$20\ m/s$ .A person standing outside on the ground also observes the ball.How will the maximum heights

$y_m$ attained and the ranges

$R$ seen by the thrower and the outside observer compare with each other?

0%
same $y_m$ different $R$
0%
same $y_m$ and $R$
0%
differently $y_m$ same $R$
0%
differently $y_m$ and $R$

Explanation

The motion of train will affect only the horizontal component of the velocity of the ball.Since , vertical component is same for both observers, the

$y_m$ will be same ,but range R will be different.

Mass

$A$ is released from rest at the top of a frictionless inclined plane

$18 m$ long and reaches the bottom

$3 s$ later. At the instant when

$A$ is released, a second mass

$B$ is projected upwards along the plate from the bottom with a certain initial velocity. Mass

$B$ travels a distance up the plane, stops and returns to the bottom so that it arrives simultaneously with

$A$ . The two masses do not collide. Initial velocity of

$B$ is

0%
$4\:ms^{-1}$
0%
$5\:ms^{-1}$
0%
$6\:ms^{-1}$
0%
$7\:ms^{-1}$

Explanation

Here, for

$A, 18=0 \times 3+\displaystyle \frac{1}{2}a \times 3^{2}$ or

$a=4 ms^{-2}$
For

$B$ , time taken to move up is given by ,

$t_1=u/a$ (

$\therefore$ the relation

$v=u+at$ here becomes

$0=u-at_{1}$ ).
Distance moved up is given by the relation

$0=u^{2}-2a S$
i.e

$S=u^{2}/2a$

For coming down the inclined plane for

$S=\displaystyle \frac{1}{2}at_{2}^{2}$

and

$t_{2}=\sqrt{\displaystyle \frac{2S}{a}}$

or

$t_{2}=\sqrt{\displaystyle \frac{2u^{2}}{a.2a}} =\displaystyle \frac{u}{a}$

But

$\displaystyle \frac{u}{a}=t_{1}$ , then

$t_1+t_2=3$ or

$\displaystyle \frac{2u}{a}=3$ or

$u=\displaystyle \frac{3a}{2}$ or

$u=\displaystyle \frac{3}{2} \times 4=6 \: ms^{-1}$

A ball is thrown vertically upwards from the ground and a student gazing out of the window sees it moving upward past him at

$\displaystyle 10 \ ms^{-1}.$ The window is at

$15$ m above the ground level. The velocity of ball

$3\ s$ after it was projected from the ground is (take

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

0%
$10 \ ms^{-1},$ up
0%
$\displaystyle 20\ ms^{-1},$ up
0%
$\displaystyle 20\ ms^{-1},$ down
0%
$\displaystyle 10\ ms^{-1},$ down

Explanation

Let the ball be thrown with initial velocity of 'u'.
After travelling 15 m, its velocity is 10 m/s.
Thus

${v}^{2}-{u}^{2}=-2gh$ (acceleration acts downwards, thus -g)
or,

${10}^{2}-{u}^{2}=-300$
giving

$u=20m/s$
After 3 sec,

$v=u-gt = 20-30 = 10 m/s$ downwards

A lift start from rest. Its acceleration is plotted against time. When it comes to rest its height above its starting point is

0%
$20\ m$
0%
$64\ m$
0%
$32\ m$
0%
$36\ m$

Explanation

From 0- 4 s the motion is with constant acceleration, the distance can be calculated from kinematics equation

$S_{0-4}=0+\frac{1}{2}2\times 4^2= 16\; m$
From 4- 8 s the motion is with constant velocity

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

$S_{4-8}=v\times t\; =8*4= 32 \; m$
From 8- 12 s the motion is with constant acceleration, the distance can be calculated from kinematics equation

$S_{8-12}=8\times 4-\frac{1}{2}2\times 4^2= 16\; m$
total distance covered

$= 16+ 32+16 m= 64 m$

A particle is projected vertically upwards and reaches the maximum height

$H$ at a time

$t=T$ . The height of the particle at any time

$t (< T)$ will be

0%
$\displaystyle g\left ( t-T \right )^{2}$
0%
$\displaystyle H-g\left ( t-T \right )^{2}$
0%
$\displaystyle \frac{1}{2}g\left ( t-T \right )^{2}$
0%
$\displaystyle H-\frac{1}{2}g\left ( T-t \right )^{2}$

Explanation

At maximum height

$,\ v=0$
Thus, from equations of motion:

$0=u-gT$ or

$u=gT$

$H=uT-\dfrac{1}{2}gT^2=gT^2-\dfrac{1}{2}gT^2$
or,

$H=0.5gT^2$
At any time

$,\ t,\ h=ut-0.5gt^2=gTt-0.5gt^2$
Add and subtract

$0.5gT^2$ , we get,

$h=(gtT-0.5gt^2-0.5gT^2)+0.5gT^2$
Put

$gT^2=H$ , we get

$h=-0.5g(T-t)^2+H=H-\dfrac{1}{2}g(T-t)^2$

A ball is dropped from a height of 49 m. The wind is blowing horizontally. Due to wind a constant horizontal acceleration is provided to the ball. Choose the correct statement (s). [Take g= 9.8

$\displaystyle ms^{-2}$ ]

0%
Path of the ball is a straight line
0%
Path of the ball is a curved one
0%
The time taken by the ball to reach the ground is 3.16 s
0%
Actual distance travelled by the ball is more than 49 m

Explanation

Vertical direction: Distance =49 m

$49 = (1/2) * 9.8 * t^{2}$

$t = 3.16 sec$

Let horizontal acceleration be

$a m/s^{2}$

$Distance_{horizontal} = 0.5 a t^{2}$

$Distance_{vertical} = 0.5*9.8 t^{2} = 4.9 t^{2}$

We see that both the distances follow the behavior

$constant * t^{2}$ .

Since at any given time

$t^{2}$ value is same for both the value of horizontal and vertical distances covered is progresses linearly with respect to each other. Hence the ball follow is straight line.

Also since there is vertical movement too, total distance traveled is more than 49 m

Answer. A, C, D.

The velocity of a particle at time

$t=0$ is

$2m/s$ . A constant acceleration of

$2m/s^{2}$ acts on the particle for

$2$ seconds at an angle of

$60$

$^{\circ}$ with its initial velocity. The magnitude of velocity and displacement of particle at the end of

$t=2s$ respectively are:

0%
$2$ $\sqrt{7}$ m/s, $4$ $\sqrt{3}$ m
0%
$2$ $\sqrt{3}$ m/s, $4$ $\sqrt{7}$ m
0%
$4$ $\sqrt{7}$ m/s, $2$ $\sqrt{3}$ m
0%
$4$ $\sqrt{3}$ m/s, $2$ $\sqrt{7}$ m

Explanation

initial velocity

$u = 2 m/s$

acceleration

$a$ =

$2 m/s$ at

$60$

$^{\circ}$

component of acceleration in direction of velocity and perpendicular to initiail velocity according to figure

${ a }_{ x }=acos{ 60 }^{ 0 }\quad =\quad 2\times \dfrac { 1 }{ 2 } =\quad 1\\ { a }_{ y }=asin{ 60 }^{ 0 }\quad =\quad 2\times \dfrac { \sqrt { 3 } }{ 2 } =\sqrt { 3 }$

${ s }_{ x }=ut+\dfrac { 1 }{ 2 } { a }_{ x }{ t }^{ 2 }=\quad 2\times 2+\dfrac { 1 }{ 2 } \times 1{ (2) }^{ 2 }=6m\quad$

${ s }_{ y }=ut+\dfrac { 1 }{ 2 } { a }_{ x }{ t }^{ 2 }=\quad 0+\dfrac { 1 }{ 2 } \times \sqrt { 3 } { (2) }^{ 2 }=2\sqrt { 3 } m$

net displacement =

$S=\sqrt { { { s }_{ x } }^{ 2 }+{ { s }_{ y } }^{ 2 } } =\quad \sqrt { 36+12 } =4\sqrt { 3 } m$

again for velocity in X any Y direction

${ v }_{ x }=u+{ a }_{ x }t=2+1\times 2=4m/s\\ { v }_{ y }=u+{ a }_{ y }t=0+\sqrt { 3 } \times 2=2\sqrt { 3 } m/s$

net velocity =

$V=\sqrt { { { v }_{ x } }^{ 2 }+{ { v }_{ y } }^{ 2 } } =\quad \sqrt { 16+12 } =2\sqrt { 7 } m/s$

thus .

option

$A$ is correct

A car moving along a straight highway with a speed of

$126 kmph$ is brought to a stop within a distance of

$200 m$ . What is the acceleration of the car and how long does it take for the car to stop?

0%
$-3.06 ms^{-2}, 11.43 s$
0%
$-6.03 ms^{-2}, 11.43 s$
0%
$-3.06 ms^{-2}, 10.34 s$
0%
$-6.03 ms^{-2}, 10.34 s$

Explanation

Initial velocity

$u = 126 km/hr$ or

$35m/s$
Final velocity

$=0 m/s$ (stopped)
Distance

$s = 200 m$
Use newton's 3rd law of motion

$v^{2} = u^{2} +2as$ ; a is acceleration or -a if it is deceleration

$0 = 35^{2} +2 \times a \times 200$

$400a = -1225$

$a = -3.0625 m/s^{2}$ ---------> This is retardation of the CAR to find time

Use first law of motion

$v = u + at$ ; first law of motion,

$0 = 35 - 3.0625t$

$t = 35/3.0625 =11.4285 sec$

A car is travelling on a straight road. The maximum velocity the car can attains is

$\displaystyle 24 ms^{-1}.$ The maximum acceleration and deceleration it can attain are

$\displaystyle 1 ms^{-2}$ and

$\displaystyle 4ms^{-2}$ respectively. The shortest time the car takes from rest to rest in a distance of 500 m is,

0%
22.4 s
0%
30 s
0%
11.2 s
0%
35.8 s

Explanation

$Displacement\quad if\quad the\quad car\quad reaches\quad max\quad speed\quad with\quad max\quad acceleration\\ { s }_{ 1 }=\dfrac { { v }^{ 2 } }{ 2a } =\dfrac { { 24 }^{ 2 } }{ 2 } =288m\\ Displacement\quad when\quad the\quad car\quad reaches\quad from\quad max\quad speed\quad to\quad zero\quad \\ { s }_{ 2 }\quad =\quad \dfrac { { u }^{ 2 } }{ 2a } =\dfrac { { 24 }^{ 2 } }{ 2\times 4 } =72\\ { s }_{ 1 }+{ s }_{ 2 }=288+72=360m\\ remaining\quad distance\quad is\quad travelled\quad at\quad maximum\quad speed,\quad for\quad which\quad the\quad \\ time\quad taken\quad is\quad \dfrac { 500-360 }{ 24 } =5.833s\\ time\quad during\quad { s }_{ 1 }=\dfrac { 24 }{ 1 } =24s\\ time\quad during\quad { s }_{ 2 }=\dfrac { 24 }{ 4 } =6s\\ total\quad time\quad is\quad 35.833s$

Starting from rest a particle is first accelerated for time

$\displaystyle t_{1}$ with constant acceleration

$\displaystyle a_{1}$ and then stops in time

$\displaystyle t_{2}$ with constant retardation

$\displaystyle a_{2}.$ Let

$\displaystyle v_{1}$ be the average velocity in this case and

$\displaystyle s_{1}$ the total displacement. In the second case

it is accelerated for the same time

$t_1$ with constant acceleration

$2a_1$ and come to rest with constant retardation

$\displaystyle a_{2}$ in time

$\displaystyle t_{3}.$ If

$\displaystyle v_{2}$ is the average velocity in this case and

$\displaystyle s_{2}$ the total displacement, then

0%
$\displaystyle v_{2}=2v_{1}$
0%
$\displaystyle 2v_{1}< v_{2}< 4v_{1}$
0%
$\displaystyle S_{2}2=S_{1}$
0%
$\displaystyle 2S_{1} < S_{2}< 4S_{1}$

Explanation

Velocity after acceleration by

$a$ for

$t_1$ time=

$a_1t_1$

It takes time

$t_2$ to retard to zero velocity, so,

$a_1t_1=a_2t_2$

$\implies t_2=\dfrac{a_1}{a_2}t_1$

Similarly for second case,

$t_3=\dfrac{2a_1}{a_2}t_1$

In first case,

Distance traveled during acceleration=

$\dfrac{1}{2}a_1t_1^2$

Distance traveled during retardation=

$\dfrac{(a_1t_1)^2}{2a_2}$

$\implies s_1=\dfrac{1}{2}a_1t_1^2+\dfrac{a_1^2t_1^2}{2a_2}$

$v_1=\dfrac{s_1}{t_1+t_2}=\dfrac{s_1}{t_1+\dfrac{a_1}{a_2}t_1}=\dfrac{1}{2}a_1t_1$

In second case,

Distance traveled during acceleration=

$a_1t_1^2$

Distance traveled during retardation=

$\dfrac{(2a_1t_1)^2}{2a_2}$

$\implies s_2=a_1t_1^2+\dfrac{2a_1^2t_1^2}{a_2}$

$v_2=\dfrac{s^2}{t_1+t_3}=\dfrac{s_2}{t_1+\dfrac{2a_1}{a_2}t_1}=a_1t_1$
Hence

$v_2=2v_1$

And

$2s_1<s_2<4s_1$

Two balls were thrown vertically upwards with different velocities, what is the shape of the graph between distance between the balls and time before either of the two collide with ground?

0%
Straight line passing through origin
0%
Parabola
0%
Circle
0%
None of the above

Explanation

Let the initial velocity of the balls be

$u_1$ and

$u_2$ (upwards) respectively.

Also the acceleration of the balls

$a_1 = -g$

$a_2 =-g$

$\therefore$ Relative acceleration of ball 1 w.r.t 2

$a_{12} =a_1 -a_2 = 0$

Initial velocity of 1 w.r.t 2

$u_{12} =u_1 - u_2$ (upwards)

Using

$S_{12} = u_{12} t - \dfrac{a_{12} t^2}{2}$

$\therefore$ Distance between them

$S_{12} = (u_1 - u_2) t$

Thus S-t graph will be a straight line having a slope

$(u_1 - u_2)$ passing through origin.

Net force acting on a particle of mass 2 kg is 10 N in north direction. At t = 0, particle was moving eastwards with 10 m/s. Find displacement and velocity of particle after 2 s.

0%
$\displaystyle 10\sqrt{5}$ m at $\displaystyle \cot^{-1}(2)$ from east towards north, $\displaystyle 10\sqrt{2}ms^{-1}\:at\:45^{\circ}$ from east towards north.
0%
$\displaystyle 10\sqrt{2}$ m at $\displaystyle \cot^{-1}(2)$ from east towards north, $\displaystyle 10\sqrt{2}ms^{-1}\:at\:45^{\circ}$ from east towards north.
0%
$\displaystyle 10\sqrt{5}$ m at $\displaystyle \cot^{-1}(2)$ from east towards north, $\displaystyle 10\sqrt{5}ms^{-1}\:at\:45^{\circ}$ from east towards north.
0%
$\displaystyle 10\sqrt{2}$ m at $\displaystyle \cot^{-1}(2)$ from east towards north, $\displaystyle 10\sqrt{5}ms^{-1}\:at\:45^{\circ}$ from east towards north.

Explanation

$a=5m/{ s }^{ 2 }along\quad north,\quad thus,\quad velocity\quad along\quad east\quad won't\quad change\\ t=2s\\ u\quad =\quad 10m/s\quad along\quad east\\ displacement\quad along\quad north\quad =\quad \frac { 1 }{ 2 } a{ t }^{ 2 }=10m\\ displacement\quad along\quad east=ut=20m\\ total\quad displacement\quad =\quad \sqrt { { 10 }^{ 2 }+{ 20 }^{ 2 } } =10\sqrt { 5 } m\\ velocity\quad along\quad north\quad =\quad at=10m/s\\ final\quad velocity\quad =\quad 10\hat { i } +10\hat { j }$

Trajectory of two particles projected from origin with speed

$\displaystyle v_{1}$ and

$\displaystyle v_{2}$ and angles

$\displaystyle \theta _{1}$ and

$\displaystyle \theta _{2}$ with positive x- axis respectively as shown in the figure given that

$\displaystyle g=-10m/s^{2}(j)$ . Choose the correct option related to diagram :

0%
$\displaystyle v_{1}-v_{2}=2v_{1}$
0%
$\displaystyle \theta_{2}-\theta _{1}=2\theta _{1}$
0%
$\displaystyle 3(v_{1}-v_{2})=0$
0%
$\displaystyle 3(\theta_{1}-\theta_{2})=-\theta _{1}$

Explanation

Horizontal range

$R = \dfrac{v^2 sin (2\theta)}{g} = \dfrac{2v^2 sin\theta cos\theta}{g}$

Maximum height attained

$H = \dfrac{v^2 sin^2 \theta}{2g}$

For particle 1 :

$R_1 = 40$ m

$H_1 =10$ m

$\therefore$

$10 = \dfrac{v_1^2 sin^2\theta_1}{2g}$ ..........(1)

And

$40 =\dfrac{2v_1^2 sin\theta_1 cos\theta_1}{g}$ .............(2)

Dividing (1) by (2), we get

$1 = tan\theta_1$

$\implies \theta_1 =45^o$

$\therefore$ From (1) we get

$10 = \dfrac{v_1^2 \times 0.5}{2 (10)}$

$\implies v_1 =20 m/s$

For particle 2 :

$R_2 = 20\sqrt{3}$ m

$H_2 =15$ m

$\therefore$

$15 = \dfrac{v_2^2 sin^2\theta_2}{2g}$ ..........(3)

And

$20\sqrt{3} =\dfrac{2v_2^2 sin\theta_2 cos\theta_2}{g}$ .............(4)

Dividing (3) by (4), we get

$\sqrt{3} = tan\theta_2$

$\implies \theta_2 =60^o$

$\therefore$ From (3) we get

$15 = \dfrac{v_2^2 \times (0.866)^2}{2 (10)}$

$\implies v_2 =20 m/s$

$\therefore$

$3(\theta_1 - \theta_2) + \theta_1 = 3 (45^o - 60^o)+ 45^o = 0$

A particle travels so that its acceleration is given by

$\vec{a}=5\,cos\,t\hat {i}-3\,sin\,t\hat {j}$ . If the particle is located at

$(-3,2)$ at time

$t=0$ and is moving with a velocity given by

$(-3\hat {i}+2\hat {j})$ . Find
(i) the velocity

$\begin{bmatrix}\vec{v}=\int \vec{a}.dt\end{bmatrix}\:time\:t$
(ii) the position vector

$[\vec{r}=\int \vec{v}.dt]$ of the particle at time

$(t>0)$ .

0%
(i) $\;\vec{v}=(5sin\;t-3)\hat {i}+(3cos\;t-1)\hat {j}$ , (ii) $\;\vec{s}=(2-5cos\;t-3t)\hat {i}+(2+3sin\;t-t)\hat {j}$
0%
(i) $\;\vec{v}=(3sin\;t-3)\hat {i}+(3cos\;t-1)\hat {j}$ , (ii) $\;\vec{s}=(2-5cos\;t-3t)\hat {i}+(2+3sin\;t-t)\hat {j}$
0%
(i) $\;\vec{v}=(4sin\;t-3)\hat {i}+(3cos\;t-1)\hat {j}$ , (ii) $\;\vec{s}=(2-5cos\;t-3t)\hat {i}+(2+3sin\;t-t)\hat {j}$
0%
None of these

Explanation

$\vec{a}=5\,cos\,t\hat {i}-3\,sin\,t\hat {j}$

$\Rightarrow \int d\vec{v}=\int 5\,cos\,t\;dt\hat {i}-\int 3\,sin\,t\;dt\hat {j}$

Therefore $\underset{-3}{\overset{v}{\int}}dv_{x}=\underset{0}{\overset{t}{\int}}5\,cos\,t\;dt\Rightarrow v_x=5\,sin\,t-3$

$\displaystyle\frac{dx}{dt}=(5sint-3)\Rightarrow \underset{-3 }{\overset{x}{\int}}dx=\underset {0}{\overset{t}{\int}}(5\,sin\,t-3)dt$

$x+3=5-5\,cos\,t-3t\Rightarrow x=2-5\,cos\,t-3t$

Similarly,

$\underset{2}{\overset{v}{\int}}dv_{y}=-\underset{0}{\overset{t}{\int}}3\,sin\,t\,dt$

$\Rightarrow v_y-2=3(cos\,t-1)\Rightarrow v_y=3\,cos\,t-1$

$\Rightarrow \underset{2}{\overset{y}{\int}}dy=\underset{0}{\overset{t}{\int}}(3\,cos\,t-1)dt$

$\Rightarrow y-2=3\;sin\,t-t\Rightarrow y=2+3\,sin\,t-t$

Thus, $\vec{v}=(5\,sin\,t-3)\hat {i}+(3\,cos\,t-1)\hat {j}$

and $\vec{s}=(2-5\,cos\,t-3\,t)\hat {i}+(2+3\,sin\,t-t)\hat {j}$

A thief is running away on a straight road with a speed of

$9 \ ms^{-1}$ . A police man chases him on a jeep moving at a speed of

$10 \ ms^{-1}$ If the instantaneous separation of the jeep from the motorcycle is 100 m, how long will it take. for the police man to catch the thief?

0%
1s
0%
19s
0%
90s
0%
100 s

A stone is thrown upwards from a tower with a velocity

$50\, ms^{-1}$ . Another stone is simultaneously thrown downwards from the same location with a velocity

$50\,ms^{-1}$ . When the first stone is at the highest point, the relative velocity of the second stone with respect to the first stone is (assume that second stone has not yet, reached the ground

0%
Zero
0%
$50\, ms^{-1}$
0%
$100\, ms^{-1}$
0%
$150\, ms^{-1}$

Explanation

Initially one stone is thrown up with a velocity of

$50ms^{-1}$
Time it takes to reach stationary.
v =0, u =50, a = -10,
v = u + at
Solving we get 5s.
Meanwhile in 5seconds the other rock initially moving down with u

$= 50ms^{-1}$ has now attained a velocity v.

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

$v = u + at = 50 + 5(10) = 100 ms^{-1}$
Hence one rock has a velocity of 0, and the other has a velocity of

$100ms^{-1}$
Hence relative velocity is

$100ms^{-1}$ option C is correct.

A ballon 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 (take

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

0%
have a displacement of 50 m
0%
cover a distance of 40 m in reaching the ground
0%
reach the ground in 4 s
0%
reach the ground in 16 s

Explanation

at t=8sec. height of ballon

$\displaystyle h=0+\frac{1}{2}\times 1.25\times (8)^{2}=40m$

and velocity of ballon

$v_{b}=0+1.25\times 8=10m/s (upward)$

And as the stone is released. It has the same initial velocity i.e 10m/sec (upward) hence it will go up by 5m and come back 5m and then fall down a distance 40m downwards so the total distance covered by stone is 40m downwards. Now if the time of journey is t then

$\displaystyle h=ut+\frac{1}{2}at^{2}\Rightarrow +40=-10t+\frac{1}{2}\times 10\times t^{2}$

$\Rightarrow t^{2}-2t-8=0\Rightarrow (t-4)(t+2)=0$

$\Rightarrow t=4s.(t=-2\;not\;possible)$

Which of the following is not an example of acceleration?

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

A particle

$X$ moving with a constant velocity

$u$ crosses a point O. At the same instant another particle

$F$ starts from rest from O with a constant acceleration

$a$ . The maximum separation between them before they meet is

0%
$u^{2}/2a$
0%
$u^{2}/a$
0%
$u/2a$
0%
$u/a$

Explanation

maximum distance will be when both of them have same velocity because after that point of time velocity of particle F will be more then that of particle X so distance between then starts decreasing.

$t=\dfrac{u}{a}$ .

distance covered by X till this time is

$S_1=u\times \dfrac{u}{a}$

distance covered by F till that time is

$s_2=\frac{1}{2}a(\dfrac{u}{a})^2= \dfrac{u^2} {2a}$

maximum distance between them

$S_1-S_2=\dfrac{u^2}{2a}$

so option A is correct.

A particle initially starts from rest, travels a distance Y in the first two seconds and a distance of X in next two seconds, if the body is moving with constant acceleration then :

0%
X = 2Y
0%
X + Y = 4X
0%
X + Y = 4Y
0%
X = 3Y

Explanation

Let the acceleration of the particle be

$a$ .

Initial speed of the particle

$u=0 m/s$

Distance travelled by it in first 2 seconds,

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

$t=2$ s

$\therefore$

$Y =0t+\dfrac{1}{2}a(2)^2$

We get

$Y = 2a$ ....(1)

Velocity of the particle after

$t=2$ s,

$v = u+ at$

$\therefore$

$v = 0+ a(2) =2a$

Distance covered by it next 2 seconds,

$X = vt+\dfrac{1}{2}at^2$ where

$t=2$ s

$\therefore$

$X = (2a)2 +\dfrac{1}{2}a(2)^2 =6 a$ .....(2)

From (1) and (2), we get

$X = 3Y$

Also

$X+Y = 3Y+Y = 4Y$

Hence options C and D are correct.

Consider a train which can accelerate with an acceleration of

$20 {cm}/{{s}^{2}}$ and slow down with deceleration of

$100 {cm}/{{s}^{2}}$ . Find the minimum time for the train to travel between the stations

$2.7 km$ apart.

0%
0
0%
90
0%
100
0%
180

An aeroplane flies along a straight line from A to B with air speed

$V$ and back again with the same air speed.If the distance between A and B is

$l$ and a steady wind blows perpendicular to AB with speed

$u$ , the total time taken for the round trip is

0%
$\dfrac{2l}{\sqrt{V^{2}-u^{2}}}$
0%
$\dfrac{2l}{\sqrt{V^{2}+u^{2}}}$
0%
$\dfrac{l}{\sqrt{V^{2}-u^{2}}}$
0%
$\dfrac{3l}{\sqrt{V^{2}-u^{2}}}$

Explanation

If the plane has to fly in straight line then he must fly at some angle theta with the AB. And component of velocity of plane in direction of air flow with respect to ground should be zero. i.e.

$u=vSin\theta$ . From here

$Sin\theta=\frac{u}{v}$

And component of velocity of plane which will help plane to cover the distance

$l$ is

$vCos\theta$

So time taken to cover distance

$l$ is given by

$\frac{l}{vCos\theta}$ and

$Cos\theta=\sqrt{1-(\frac{u}{v})^2}$ .

So time taken from A to B is

$\frac{l}{\sqrt{v^2-u^2}}$ same will be case while returning so total time will be

$\frac{2l}{\sqrt{v^2-u^2}}$

So best option is A.

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

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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