Kinematics-1D

Slope of $ s - t $ graph = Velocity = Positive
At $ t = 0 $, $ s \neq 0 $. Further at $ t = t_0 $; $ s = 0 $, $ v \neq 0 $.
Thus, v is constant.

By differentiating $a - t$ equation two times, we will get $s - t$ equation.
Further

Ascending means velocity is in upward direction. Speed is decreasing. It means acceleration is downwards.
Further, the body moves in the direction of velocity.

From the given x-t graph, it is clear that
at $t = \Omega ,\frac{dx}{dt} = 0$
∴ $ V_{velocity}, v = \frac{dx}{dt} = \Omega $ at t = 0
As time passes $\frac{dx}{dt}$ increases and then decreases.
∴ Velocity and hence, acceleration changes.

$x = 4(t - 2) + a(t - 2)^{2}$
At $t = \Omega, x = -8 + 4a = 4a - 8$
$v = \frac {d x}{d t} = 4 + 2 a (t - 2)$
At $t = 0, v = 4 - 4a = 4(1 - a)$
But acceleration, $a = \frac{d^2x}{dt^2} = 2a$

For maximum and minimum displacement, we have to keep in mind the magnitude and direction of maximum velocity.
As maximum velocity in positive direction is $ v_{0} $ maximum velocity in opposite direction is also $ v_{0} $ .
Maximum displacement in one direction $= v_{0}T$
Maximum displacement in opposite directions $= -v_{0}T$
Hence, $-v_{0}T < x < v_{0}T$.

Distance travelled in last second of upward journey is independent of velocity of projection.
Distance travelled $= \frac{1}{2} g t^2 = \frac{1}{2}\times 9.8\times (10^{2} = 4.9\mathrm{m}$

Average velocity $= \frac{\text{Displacement}}{\text{Time}}$
$= \frac {\text { Area of } v - t \text { graph }}{\text { Time }}$
$= \frac {(1 / 2) v _ {m} t _ {0}}{t _ {0}} = \frac {1}{2} v _ {m}$

$2 2. 5 = 0 + \frac {1}{2} \times 9. 8 \times t ^ {2} = 4. 9 t ^ {2}$
$\Rightarrow \quad t = \sqrt {\frac {1 2 2 . 5}{4 . 9}} = \sqrt {2 5} = 5 \mathrm{s}$
For second ball, $122.5 = u(5 - 2) + \frac{1}{2}\times 9.8\times (5 - 2)^{2}$
= 3 u + 4 4. 1
$\Rightarrow \quad 3 u = 1 2 2. 5 - 4 4. 1$

Given, speed of river, $ v_{R} = 10 \, ms^{-1} $
Speed of swimmer in still water, $ v_{SN} = 20 \, ms^{-1} $

For the shortest path to cross the river, he should swim at an angle $ (90^{\circ} + \theta) $ with the stream flow. From the above figure,
$\mathbf {v} _ {\mathrm{SN}} = \mathbf {v} _ {\mathrm{R}} + \mathbf {v} _ {\mathrm{S}}$
So, angle $\theta$ is given by
$\sin \theta = \left| \frac {\mathbf {v} _ {R}}{\mathbf {v} _ {S N}} \right| = \frac {1 0}{2 0} = \frac {1}{2} \Rightarrow \theta = 3 0 ^ {\circ}$
As the river is flowing in east direction, so the shoal wind towards west.

$\because$ Runner starts from $O$ and goes to $O$ following the path $OQRO$, so net displacement is zero.
$\begin{array}{l} \because \text { Average speed } = \frac {\text { Total distance }}{\text { Total time }} = \frac {O Q + Q R + R O}{\text { Total time }} \\ = \frac {1 \mathrm{km} + (2 \pi r) (l / 4) + 1 \mathrm{km}}{1 \mathrm{h}} \\ = 2 + \frac {\pi}{2} = 3 5 7 \mathrm{km} ^ {\mathrm{h} - 1} \\ \end{array}$

For the given $v$-$x$ graph,
Slope $= -v_{0} / x_{0}$
and intercept $= v_{0}$
From the general equation of straight line, i.e.
y = m x + c
where, m is slope and c is the intercept.
The equation of motion of the v-x graph can be given as
$v = - \frac {v _ {0} x}{x _ {0}} + v _ {0} \quad \dots \tag {i}$
(-ve sign signifies that the slope is decreasing)
On differentiating Eq. (ii) with respect to $ t $, we get
$\frac {d v}{d t} = \frac {v _ {0}}{x _ {0}} \frac {d x}{d t} + 0 \text { or } a = \frac {v _ {0}}{x _ {0}} v \tag {11}$
Putting the value of $ v $ from Eq. (i) in Eq. (ii), we get
$a = - \frac {v _ {0}}{x _ {0}} \left[ - \frac {v _ {0}}{x _ {0}} x + v _ {0} \right] = \frac {v _ {0} ^ {2}}{x _ {0} ^ {2}} x - \frac {v _ {0} ^ {2}}{x _ {0} ^ {2}}$
This equation shows that at x = 0,
$a = - \frac {v _ {0} ^ {2}}{x _ {0}} = \text { constant and } a = 0, x = x _ {0}.$
So, option (c) is correct.