Motion

From $t=0$ to $t=2 \mathrm{~s}, a=+2 \mathrm{~m} / \mathrm{s}^2$

$ \therefore \quad v=a t=2 t $

or $v-t$ graph is a straight line passing through origin with slope $2 \mathrm{~m} / \mathrm{s}^2$.

At the end of 2 s ,

$ v=2 \times 2=4 \mathrm{~m} / \mathrm{s} $

From $t=2$ to $4 \mathrm{~s}, a=0$

Hence, $v=4 \mathrm{~m} / \mathrm{s}$ will remain constant.

From $t=4 \mathrm{~s}$ to $6 \mathrm{~s}, a=-4 \mathrm{~m} / \mathrm{s}^2$

Hence, $v=u-a t=4-4 t$ (with $t=0$ at 4 s )

$v=0$ at $t=1 \mathrm{~s}$ or at 5 s from origin.

At the end of $6 \mathrm{~s}($ or $t=2 \mathrm{~s}), v=-4 \mathrm{~m} / \mathrm{s}$.

$ \text { Corresponding } v-t \text { graph is as shown below } $

When a projectile is launched with a velocity of $ 40 \, \text{m/s} $ at an angle $\theta$ to the horizontal, and its inclination with the horizontal becomes zero after $ t $ seconds, we want to determine the value of $\cot \theta$.

Maximum Height Condition: The projectile's inclination with the horizontal becomes zero at its maximum height.

Time to Reach Maximum Height:

$ t = \frac{u \sin \theta}{g} $

Solving for $ u $, we get:

$ u = \frac{g t}{\sin \theta} \quad \ldots \text{(i)} $

Horizontal Range $ R $:

$ R = u \cos \theta \times (2t) $

Solving for $\cos \theta$:

$ \cos \theta = \frac{R}{2u t} $

Substituting $ u $ from (i):

$ \cos \theta = \frac{R}{2 \cdot \frac{g t}{\sin \theta} \cdot t} $

Simplifying gives:

$ \cos \theta = \frac{R \sin \theta}{2 g t^2} $

Finding $\cot \theta$:

$ \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{R}{2 g t^2} $

Substituting $ g = 10 \, \text{m/s}^2 $:

$ \cot \theta = \frac{R}{2 \times 10 t^2} = \frac{R}{20 t^2} $

In the following figure, we have

$v_B \sin \theta=v_w$

$\sin \theta=\frac{v_w}{v_B}=\frac{1}{\sqrt{2}} \Rightarrow \theta=45^{\circ} \quad\left(\because v_B=\sqrt{2} v_w\right)$

Time to cross the river,

$t=\frac{D}{v_B \cos \theta}=\frac{D}{v_B \cos 45^{\circ}}=\frac{\sqrt{2} D}{v_B}$

Given, $v = 3\widehat i + 4\widehat j$

$\therefore$ $v = |v| = \sqrt {{3^2} + {4^2}} $

$\Rightarrow$ $v = 5$ m/s

From figure,

$\sin \theta = {4 \over 5}$ and $\cos \theta = {3 \over 5}$

$\because$ $R = {{{v^2}\sin 2\theta } \over g}$

$ = {{{v^2}\,.\,2\sin \theta \,.\,\cos \theta } \over g}$

$ = {{5 \times 5 \times 2 \times {4 \over 5} \times {3 \over 5}} \over {10}} = 2.4$ m

Distance moved by man in 5 min with velocity of 45 m/s is

Distance = Speed $\times$ Time

45 m/s $\times$ (5 $\times$ 60) s = 13500 m

When he move back, it covers the same distance to come back to its original position.

Now, time taken = 1.5 min

= 1.5 $\times$ 60 s = 90 s

Distance travelled = 13500 m

$\therefore$ Speed $ = {{distance} \over {Time}} = {{13500} \over {90}} = 150$ m/s

According to equation of motion, in vector form,

$v = u - gt\,.\,\widehat j$

$ \Rightarrow v = \widehat i + \widehat j - 10t\,\widehat j$

Integrating w.r.t. 't' both sides, we get

$r = \int {v\,dt = \int {(\widehat i + \widehat j - 10t\,\widehat j)\,dt} } $

or, $x\widehat i + y\widehat j = t\widehat i + t\widehat j - 5{t^2}\widehat j$

$\therefore$ On comparing, we get

$x = t$

and $y = t - 5{t^2}$

$ \Rightarrow y = x - 5{x^2}$

This is the required equation of trajectory of particle.

Given:

At time $ t = 0 $, the particle starts from the origin $ (0, 0) $ with a speed of $ v = 1\, \text{m/s} $.

The particle moves along the path defined by $ y = \frac{x^2}{2} $.

Understanding the Particle's Motion

Since the particle follows the path $ y = \frac{x^2}{2} $, we can find relationships between its velocity and acceleration components.

1. Velocity Components

Let $ x(t) $ and $ y(t) $ be the particle's position as functions of time. Then:

$ y = \frac{x^2}{2} $

Differentiate both sides with respect to time $ t $:

$ \frac{dy}{dt} = x \frac{dx}{dt} $

The velocity components are:

$ v_x = \frac{dx}{dt} $

$ v_y = \frac{dy}{dt} = x v_x $

2. Acceleration Components

The acceleration components are:

$ a_x = \frac{dv_x}{dt} $

$ a_y = \frac{dv_y}{dt} = \frac{d}{dt}(x v_x) = v_x^2 + x a_x $

Analyzing Each Option

Option A

Statement: $ a_x = 1\, \text{m/s}^2 $ implies that when the particle is at the origin, $ a_y = 1\, \text{m/s}^2 $.

Verification:

At the origin $ x = 0 $, and from initial conditions, $ v_x = 1\, \text{m/s} $.

Using the expression for $ a_y $:

$ a_y = v_x^2 + x a_x = (1\, \text{m/s})^2 + 0 \times a_x = 1\, \text{m/s}^2 $

Conclusion: Option A is true.

Option B

Statement: $ a_x = 0 $ implies $ a_y = 1\, \text{m/s}^2 $ at all times.

Verification:

If $ a_x = 0 $, then $ v_x = \text{constant} = 1\, \text{m/s} $.

Position along $ x $: $ x = v_x t = t $.

Then $ v_y = x v_x = t \times 1\, \text{m/s} = t\, \text{m/s} $.

Therefore, $ a_y = \frac{dv_y}{dt} = 1\, \text{m/s}^2 $.

Conclusion: Option B is true.

Option C

Statement: At $ t = 0 $, the particle's velocity points in the $ x $-direction.

Verification:

The slope of the path is $ \frac{dy}{dx} = x $.

At $ x = 0 $, $ \frac{dy}{dx} = 0 $, meaning the path is horizontal.

Thus, $ v_y = 0 $ at $ t = 0 $, and $ v_x = 1\, \text{m/s} $.

Conclusion: Option C is true.

Option D

Statement: $ a_x = 0 $ implies that at $ t = 1\, \text{s} $, the angle between the particle's velocity and the $ x $-axis is $ 45^\circ $.

Verification:

With $ a_x = 0 $, $ v_x = 1\, \text{m/s} $.

At $ t = 1\, \text{s} $, $ x = 1\, \text{m} $.

Then $ v_y = x v_x = 1\, \text{m} \times 1\, \text{m/s} = 1\, \text{m/s} $.

The angle $ \theta $ between $ \vec{v} $ and the $ x $-axis is:

$ \tan \theta = \frac{v_y}{v_x} = \frac{1\, \text{m/s}}{1\, \text{m/s}} = 1 \implies \theta = 45^\circ $

Conclusion: Option D is true.

Final Answer

All options A, B, C, and D are true.

Distance = area under speed-time graph as given in question.

$\therefore$ Distance travelled in 20s = area of $\Delta$OAB + area of $\Delta$BCD

$ = {1 \over 2} \times 6 \times 10 + {1 \over 2} \times 6 \times 10 = 60$ m

Displacement in 20 s = area of $\Delta$OAB $-$ area of $\Delta$BCD $ = {1 \over 2} \times 6 \times 10 - {1 \over 2} \times 6 \times 10 = 30 - 30 = 0$

Hence, average velocity $ = {{displacement} \over {time}} = {0 \over t} = 0$

Acceleration, $a = {{dv} \over {dt}} \Rightarrow dv = adt$

Integrating both sides, $\int {dv = \int {adt} } $

Given, $a = (3{t^2} - 2t + 1)$ ms^$-$2

Substituting this value in above equation, we get

$v = \int {(3{t^2} - 2t + 1)dt} $

$ \Rightarrow v = {t^3} - {t^2} + t + C$ ...... (i)

At t = 1s, v = 1 m/s

Substituting these values in Eq. (i), we get

$\because$ $1 = 1 - 1 + 1 + C \Rightarrow C = 0$

Substituting this value of C in Eq. (i), we get

$ \Rightarrow v = {t^3} - {t^2} + t$

when t = 4s, $v = {(4)^2} - {(4)^2} + 4 = 52$ m/s

From figure we can write

$\overrightarrow P = b\overrightarrow R = \overrightarrow S $

It is given that $\overrightarrow R = \overrightarrow Q - \overrightarrow P $. Put in above equation we get

$\overrightarrow P + b(\overrightarrow Q - \overrightarrow P ) = \overrightarrow S \Rightarrow \overrightarrow P + b\overrightarrow Q - b\overrightarrow P = \overrightarrow S $

$ \Rightarrow \overrightarrow S = \overrightarrow P (1 - \overrightarrow b ) + b\overrightarrow Q $

Density $\rho = {{mass\,(m)} \over {volume\,(V)}}$

Volume of sphere $ = {4 \over 3}\pi {R^3} \Rightarrow \rho = {m \over {{4 \over 3}\pi {R^3}}} = {{3m} \over {4\pi {R^3}}}$

Rearranging it we get

$\rho {R^2} = {3 \over {4\pi }}m$

It is given that mas m remains constant. Let the constant be k. Therefore,

$\rho {R^3} = k$

Differentiating it w.r.t. time, we get

${d \over {dt}}(\rho {R^3}) = 0$ (Since, differentiation of constant is zero)

${R^3}{{d\rho } \over {dt}} + 3{R^2}\rho {{dR} \over {dt}} = 0$

Now ${{dR} \over {dt}}$, that is, rate of change of radius is equal to velocity. Therefore,

${R^3}{{d\rho } \over {dt}} + 3{R^2}\rho v = 0 \Rightarrow 3{R^2}\rho v = - {R^3}{{d\rho } \over {dt}}$

$ \Rightarrow v = {{ - 1} \over 3}{{{R^3}} \over {{R^2}}}{1 \over \rho }{{d\rho } \over {dt}} = {{ - 1R} \over 3}{1 \over \rho }{{d\rho } \over {dt}}$

It is given that rate of fractional change in density $\left( {{1 \over \rho }{{d\rho } \over {dt}}} \right)$ is constant. Thus,

v $\propto$ R

Thus velocity of any point on surface of expanding sphere is proportional to R.

Statement 1 is True.

For a moving observer, the nearly objects appear to move in the opposite direction at a large speed. This is because the angular speed of the nearby object w.r.t to observer is large. As the object moves away, the angular velocity decreases and therefore its speed seems to be less. The distant object almost remains stationary.

Statement 2 is True but related to relative velocity

${\overline V _{21}} = {\overline V _{2G}} - {\overline V _{1G}}$

Where, G = Laboratory frame

Thus, both the statement are true but statement 2 is not the correct explanation of statement 1.

Also condition for equilibrium of the piston is

Mg = (P$_0$ $-$ P)$\pi$R$^2$

$ \Rightarrow P = {{ - Mg} \over {\pi {R^2}}} + {P_0}$

Cylinder is thermally conducting, the temperature remains the same.

$ \Rightarrow {P_0} \times (2L \times \pi {R^2}) = Py \times \pi {R^2}$

$ \Rightarrow y = {{{P_0}} \over P} \times 2L$

$ = {{{P_0}} \over {\left[ {{P_0} - {{Mg} \over {\pi {R^2}}}} \right]}} \times 2L$

$ = \left[ {{{{P_0} \times \pi {R^2}} \over {{P_0}\pi {R^2} - Mg}}} \right] \times 2L$