Motion in a Straight Line — Page 2 | JEE Physics

Q11

Two buses P and Q start from a point at the same time and move in a straight line and their positions are represented by

{X_P}(t) = \alpha t + \beta {t^2}

and

{X_Q}(t) = ft - {t^2}

. At what time, both the buses have same velocity?

A

{{\alpha - f} \over {1 + \beta }}

B

{{\alpha + f} \over {2(\beta - 1)}}

C

{{\alpha + f} \over {2(1 + \beta )}}

D

{{f - \alpha } \over {2(1 + \beta )}}

Correct Answer

Option D

Solution

{X_P} = \alpha t + \beta {t^2}

{X_Q} = ft - {t^2}

$\therefore$

{V_P} = \alpha + 2\beta t

{V_Q} = f - 2t

$\because$

{V_P} = {V_Q}

\Rightarrow \alpha + 2\beta t = f - 2t

\Rightarrow t = {{f - \alpha } \over {2(1 + \beta )}}

Q12

A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height

\dfrac{h}{3}

while going up and coming down respectively.

A

\dfrac{\sqrt{2}-1}{\sqrt{2}+1}

B

\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}

C

\dfrac{\sqrt{3}-1}{\sqrt{3}+1}

D

\dfrac{1}{3}

Correct Answer

Option B

Solution

A ball is thrown vertically upward with a certain velocity, reaching a maximum height $h$ .

We need to find the ratio of the times it is at height $\dfrac{h}{3}$ while ascending and descending, respectively.

The initial velocity of the ball $v$ can be given by: $v = \sqrt{2gh}$ When the ball is at height $\dfrac{h}{3}$ , the equation of motion is: $\dfrac{h}{3} = \sqrt{2gh} \cdot t - \dfrac{1}{2}gt^2$ Rearranging the equation, we get a quadratic equation in terms of $t$ : $\dfrac{g}{2} t^2 - \sqrt{2gh} \cdot t + \dfrac{h}{3} = 0$ The ratio of the times taken to ascend and descend to the height $\dfrac{h}{3}$ can be found using the quadratic formula solution for $t$ , considering the corresponding velocities: $\dfrac{t_1}{t_2} = \dfrac{\sqrt{2gh} + \sqrt{2gh - \dfrac{2gh}{3}} }{\sqrt{2gh} - \sqrt{2gh - \dfrac{2gh}{3}}}$ Simplifying the terms inside the fraction: $= \dfrac{\sqrt{2} + \dfrac{2}{\sqrt{3}}}{\sqrt{2} - \dfrac{2}{\sqrt{3}} } = \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$

Q13

A juggler throws balls vertically upwards with same initial velocity in air. When the first ball reaches its highest position, he throws the next ball. Assuming the juggler throws n balls per second, the maximum height the balls can reach is

A g/2n

B g/n

C 2gn

D g/2n2

Correct Answer

Option D

Solution

The juggler throws n balls per second. $\therefore$ Interval between balls =

{1 \over n}

seconds At maximum height velocity of first ball

v = 0

Juggler throws all balls with same initial velocity

= u

For 1st ball,

{u_1} = u

,

{v_1} = 0

,

a = - g

,

S = {H_{\max }}

Using formula,

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

0 = {u^2} - 2g\,{H_{\max }}

\Rightarrow {H_{\max }} = {{{u^2}} \over {2g}}

...... (1) And using formula,

v = u + at

\Rightarrow 0 = u - gt

\Rightarrow t = {u \over g}

$\therefore$ Time taken by ball 1 to reach maximum height

({H_{\max }}) = {u \over g}

According to the question,

t = {1 \over n}

\Rightarrow {u \over g} = {1 \over n}

\Rightarrow u = {g \over n}

....... (2) Putting value of

u

in equation (1), we get

{H_{\max }} = {{{{\left( {{g \over n}} \right)}^2}} \over {2g}}

= {{{g^2}} \over {2{n^2}g}}

= {g \over {2{n^2}}}

Q14

A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity 'v' that varies as

v = \alpha \sqrt x

. The displacement of the particle varies with time as

A t2

B t

C t1/2

D t3

Correct Answer

Option A

Solution

Given the velocity function:

v = \alpha \sqrt{x}

We know that velocity $v$ is the rate of change of displacement with respect to time:

\therefore \frac{dx}{dt} = \alpha \sqrt{x}

Rearranging and separating variables, we get:

\Rightarrow \frac{dx}{\sqrt{x}} = \alpha \, dt

To solve this, we integrate both sides:

\int\limits_{0}^{x} \frac{dx}{\sqrt{x}} = \alpha \int\limits_{0}^{t} dt

This gives us:

\Rightarrow \left[ \frac{2\sqrt{x}}{1} \right]_{0}^{x} = \alpha \left[t\right]_{0}^{t}

Simplifying, we obtain:

\Rightarrow 2\sqrt{x} = \alpha t

Squaring both sides:

\Rightarrow x = \frac{\alpha^2}{4} t^2

Thus, the displacement $x$ varies with time $t$ as:

x \propto t^2

Q15

Two trains 'A' and 'B' of length '

l

' and '

4 l

' are travelling into a tunnel of length '

\mathrm{L}

' in parallel tracks from opposite directions with velocities

108 \mathrm{~km} / \mathrm{h}

and

72 \mathrm{~km} / \mathrm{h}

, respectively. If train 'A' takes

35 \mathrm{~s}

less time than train 'B' to cross the tunnel then. length '

L

' of tunnel is : (Given

\mathrm{L}=60 l

)

A 900 m

B 1200 m

C 1800 m

D 2700 m

Correct Answer

Option C

Solution

Let's start by converting the velocities of both trains to m/s: Train A:

108 \frac{km}{h} \times \frac{1000 m}{km} \times \frac{1 h}{3600 s} = 30 \frac{m}{s}

Train B:

72 \frac{km}{h} \times \frac{1000 m}{km} \times \frac{1 h}{3600 s} = 20 \frac{m}{s}

To cross the tunnel, Train A has to cover a distance equal to the length of the tunnel plus its own length:

L + l

Similarly, Train B has to cover a distance equal to the length of the tunnel plus its own length:

L + 4l

We are given that Train A takes 35 seconds less time than Train B to cross the tunnel.

Let's denote the time taken by Train A as

t_A

and the time taken by Train B as

t_B

. Then, we have:

t_B = t_A + 35

Using the formula distance = velocity × time, we can write the equations for both trains: Train A:

(L + l) = 30t_A

Train B:

(L + 4l) = 20t_B

Now, we can substitute

t_B = t_A + 35

in the equation for Train B:

(L + 4l) = 20(t_A + 35)

We have two equations and two unknowns (

L

and

t_A

). We can solve this system of equations by eliminating one of the unknowns. Let's eliminate

t_A

by expressing it from the equation for Train A:

t_A = \frac{L + l}{30}

Now, substitute this expression for

t_A

in the equation for Train B:

(L + 4l) = 20\left(\frac{L + l}{30} + 35\right)

Multiplying both sides by 30 to get rid of the fraction:

30(L + 4l) = 20(L + l) + 20 \times 35 \times 30

Expanding the equation:

30L + 120l = 20L + 20l + 21,000

Simplify:

10L + 100l = 21,000

Since we are given that

L = 60l

, substitute this into the equation:

10(60l) + 100l = 21,000

Solve for

l

:

700l = 21,000

l = 30

Now, substitute the value of

l

back into the equation for

L

:

L = 60l = 60 \times 30 = 1800

So, the length of the tunnel is: 1800 m

Q16

A ball is thrown vertically upward with an initial velocity of

150 \mathrm{~m} / \mathrm{s}

. The ratio of velocity after

3 \mathrm{~s}

and

5 \mathrm{~s}

is

\dfrac{x+1}{x}

. The value of

x

is ___________.

\left\{\right.

take,

\left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right\}

A

-5

B 10

C 5

D 6

Correct Answer

Option C

Solution

To solve this problem, we can use the following equation of motion for the vertical velocity at any given time

t

:

v = u - gt

Where: -

v

is the final velocity at time

t

-

u

is the initial velocity (150 m/s) -

g

is the acceleration due to gravity (10 m/s²) -

t

is the time in seconds First, we need to find the velocities at

t = 3 \mathrm{~s}

and

t = 5 \mathrm{~s}

. For

t = 3 \mathrm{~s}

:

v_3 = 150 - (10)(3) = 150 - 30 = 120 \mathrm{~m} / \mathrm{s}

For

t = 5 \mathrm{~s}

:

v_5 = 150 - (10)(5) = 150 - 50 = 100 \mathrm{~m} / \mathrm{s}

Now we need to find the ratio of these velocities:

\frac{v_3}{v_5} = \frac{120}{100} = \frac{6}{5} = \frac{x + 1}{x}

Next, we can set up an equation to find the value of

x

:

\frac{6}{5} = \frac{x + 1}{x}

Now, we can solve for

x

:

6x = 5(x + 1)

6x = 5x + 5

x = 5

The value of

x

is 5.

Q17

Given below are two statements Statement I : Area under velocity- time graph gives the distance travelled by the body in a given time. Statement II : Area under acceleration- time graph is equal to the change in velocity- in the given time. In the light of given statements, choose the correct answer from the options given below.

A Both Statement I and Statement II are False.

B Both Statement I and Statement II are true.

C Statement I is incorrect but Statement II is true.

D Statement I is correct but Statement II is false.

Correct Answer

Option C

Solution

Area under velocity time graph gives displacement of body in given time.

Area under acceleration time graph gives change in velocity in the given time.

So Statement I false but Statement II True

Q18

A body starts moving from rest with constant acceleration covers displacement

S_1

in first

(p-1)

seconds and

\mathrm{S}_2

in first

p

seconds. The displacement

\mathrm{S}_1+\mathrm{S}_2

will be made in time :

A

(2 p+1) s

B

(2 p-1) s

C

\left(2 p^2-2 p+1\right) s

D

\sqrt{\left(2 p^2-2 p+1\right)} s

Correct Answer

Option D

Solution

Let's denote the constant acceleration with which the body moves as

a

. We know that the displacement

S

covered by a body starting from rest under a constant acceleration

a

in time

t

is given by the equation of motion:

S = \frac{1}{2} a t^2

. Considering the displacement

\mathrm{S}_1

in first

(p-1)

seconds, we apply the equation of motion:

\mathrm{S}_1 = \frac{1}{2} a (p-1)^2

Similarly, for the displacement

\mathrm{S}_2

in first

p

seconds:

\mathrm{S}_2 = \frac{1}{2} a p^2

To find out the total time it will take to cover the displacement

\mathrm{S}_1+\mathrm{S}_2

, we first find the sum of these two displacements:

\mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a (p-1)^2 + \frac{1}{2} a p^2

Let's simplify this:

\mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a \left((p-1)^2 + p^2\right)

\mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a \left(p^2 - 2p + 1 + p^2\right)

\mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a \left(2p^2 - 2p + 1\right)

\mathrm{S}_1+\mathrm{S}_2 = a \left(\frac{2p^2 - 2p + 1}{2}\right)

If we consider the total displacement

\mathrm{S}_1+\mathrm{S}_2

is to be covered in a time

t

seconds from rest, we should set this equal to the equation of motion:

\mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a t^2

Equating the two equations:

a \left(\frac{2p^2 - 2p + 1}{2}\right) = \frac{1}{2} a t^2

Since

a \neq 0

, we can simplify by dividing both sides by

\frac{1}{2} a

:

\frac{2p^2 - 2p + 1}{2} = \frac{t^2}{2}

2p^2 - 2p + 1 = t^2

To find

t

, we take the square root of both sides:

t = \sqrt{2p^2 - 2p + 1}

Therefore, the time taken to cover the displacement

\mathrm{S}_1+\mathrm{S}_2

will be:

t = \sqrt{(2p^2 - 2p + 1)}\ s

Hence, the correct option would be: Option D:

\sqrt{(2p^2 - 2p + 1)}\ s

Q19

A particle is moving in a straight line. The variation of position '

x

' as a function of time '

t

' is given as

x=\left(t^3-6 t^2+20 t+15\right) m

. The velocity of the body when its acceleration becomes zero is :

A 6 m/s

B 10 m/s

C 8 m/s

D 4 m/s

Correct Answer

Option C

Solution

The position equation is given by:

x = t^3 - 6t^2 + 20t + 15

First, compute the velocity

v

by differentiating the position function with respect to time:

\frac{d x}{d t} = v = 3t^2 - 12t + 20

Next, compute the acceleration

a

by differentiating the velocity function with respect to time:

\frac{d v}{d t} = a = 6t - 12

We need to find the time

t

when the acceleration is zero:

6t - 12 = 0

t = 2 \, \mathrm{sec}

Now, find the velocity at

t = 2 \, \mathrm{sec}

:

v = 3(2)^2 - 12(2) + 20

v = 8 \, \mathrm{m/s}

Q20

A particle moves along the

x

-axis and has its displacement

x

varying with time t according to the equation:

x=\mathrm{c}_0\left(\mathrm{t}^2-2\right)+\mathrm{c}(\mathrm{t}-2)^2

where

\mathrm{c}_0

and c are constants of appropriate dimensions. Then, which of the following statements is correct?

A the acceleration of the particle is

2\left(c+c_0\right)

B the acceleration of the particle is

2 c_0

C the acceleration of the particle is 2 c

D the initial velocity of the particle is

4 c

Correct Answer

Option A

Solution

To determine the acceleration of the particle, we start by differentiating the displacement function with respect to time $t$ .

The displacement of the particle is given by: $x = c_0(t^2 - 2) + c(t - 2)^2$ First, compute the velocity $v$ by differentiating $x$ with respect to $t$ : $v = \dfrac{dx}{dt} = \dfrac{d}{dt}[c_0(t^2 - 2) + c(t - 2)^2]$ This gives: $v = 2c_0 t + 2c(t - 2)$ Next, determine the acceleration $a$ by differentiating the velocity function with respect to time $t$ : $a = \dfrac{dv}{dt} = \dfrac{d}{dt}[2c_0 t + 2c(t - 2)]$ Calculating this gives: $a = 2c_0 + 2c$ Hence, the acceleration of the particle is $2c_0 + 2c$ .