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.

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$

When a projectile is fired from the ground, it follows a parabolic path because the force of gravity acts on it, causing it to accelerate downward. At the top of the projectile's path, its vertical velocity becomes zero, while the horizontal velocity remains constant. At this point, the speed of the projectile is minimum because it is only moving horizontally and has no vertical component of velocity.

As the projectile continues to fall, its speed increases because of the acceleration due to gravity. When the projectile reaches the ground, its speed will be at a maximum, assuming there is no air resistance.

Therefore, the statement that the speed of the projectile is minimum at the top of its path is true.

Acceleration due to gravity is independent of the mass of the body.