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