Kinematics-1D

$2 2. 5 = 0 + \frac {1}{2} \times 9. 8 \times t ^ {2} = 4. 9 t ^ {2}$
$\Rightarrow \quad t = \sqrt {\frac {1 2 2 . 5}{4 . 9}} = \sqrt {2 5} = 5 \mathrm{s}$
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.

D

$s = ut - \frac{1}{2} gt^2$
$\begin{array}{l} {= 1 6 - \frac {1}{2} \times 1 0 \times 1 6} \\ {= - 6 4 m} \end{array}$
Height of bridge above water surface = 64 m.

Between time interval 20 sec to 40 sec, there is non-zero acceleration and retardation. Hence distance travelled during this interval
= Area between time interval 20 sec to 40 sec
$= \frac {1}{2} \times 2 0 \times 3 + 2 0 \times 1 = 3 0 + 2 0 = 5 0 m.$

As in distance vs time graph slope is equal to speed in the given graph slope increase initially which is incorrect.

Slope of v-t graph gives acceleration
$v ^ {2} = 0 + 2 \left(\frac {q E}{m}\right) x \Rightarrow v ^ {2} = \frac {2 q E}{m} x.$
$\Rightarrow$ Acceleration will be

When body going vertically upwards, reaches at the highest point, then it is momentarily at rest and it then reverses its direction. At the highest point of motion, its velocity is zero but its acceleration is equal to acceleration due to gravity.

As motion is governed by force of gravity and acceleration due to gravity (g) is independent of mass of object.

When a body falling freely, only gravitational force acts on it in vertically downward direction. Due to this downward acceleration the velocity of a body increases and will be maximum when the body touches the ground.

Since velocity is a vector quantity, hence as its direction changes keeping magnitude constant, velocity is said to be changed. But for constant speed in equal time interval distance travelled should be equal.

For first projectile, $h_1 = ut - \frac{1}{2} gt^2$ For second projectile, $h_2 = u(t - T) - \frac{1}{2} g(t - T)^2$ When both meet i.e. $h_1 = h_2$ $ut - \frac{1}{2} gt^2 = u(t - T) - \frac{1}{2} g(t - T)^2$ $\Rightarrow uT + \frac{1}{2} gT^2 = gtT \Rightarrow t = \frac{u}{g} + \frac{T}{2}$ and $h_1 = u\left(\frac{u}{g} + \frac{T}{2}\right) - \frac{1}{2} g\left(\frac{u}{g} + \frac{T}{2}\right)^2 = \frac{u^2}{2g} - \frac{gT^2}{8}$ .

Average velocity = $\frac{Total displacement}{Time taken}$ = $\frac{25}{75/15}$ = 5m/s

$v_{av} = \frac{Total\ distance}{Time\ taken} = \frac{x}{\frac{x/3}{v} + \frac{x/3}{2v} + \frac{x/3}{3v}} = \frac{18}{11}v$

Let height of tower is h and body takes t time to reach to ground when it fall freely.
∴ $h = \frac{1}{2} g t^{2}$ ...(i)
In last second i.e. $t^{th}$ sec body travels = 0.36 h
It means in rest of the time i.e. in $(t-1)$ sec it travels
= h - 0.36 h = 0.64 h
Now applying equation of motion for $(t-1)$ sec $0.64 h = \frac{1}{2} g (t-1)^{2}$ ...(ii)
From (i) and (ii) we get, t = 5 sec and h = 125m

Average acceleration = $\frac{Change in velocity}{Time taken} = \frac{v_{2} - v_{1}}{t_{2} - t_{1}}$ $= \frac{\left[10 + 2(5)^{2}\right] - \left[10 + 2(2)^{2}\right]}{3} = \frac{60 - 18}{3} = 14 \, m/s^{2}$ .

As $v^2 = u^2 - 2as \Rightarrow u^2 = 2as \quad (\because v = 0)$ $\Rightarrow u^2 \propto s \Rightarrow \frac{u_2}{u_1} = \left(\frac{s_2}{s_1}\right)^{1/2}$ $\Rightarrow u_2 = \left(\frac{9}{4}\right)^{1/2} u_1 = \frac{3}{2} u_1 = 300cm/s$ .

Let car B catches, car A after 't' sec, then $60t + 2.5 = 70t - \frac{1}{2} \times 20 \times t^{2}$ $\Rightarrow 10t^{2} - 10t + 2.5 = 0 \Rightarrow t^{2} - t + 0.25 = 0$ $\therefore t = \frac{1 \pm \sqrt{1 - 4 \times (0.25)}}{2} = \frac{1}{2} hr$

As $v^{2}=u^{2}+2as\Rightarrow(2u)^{2}=u^{2}+2as\Rightarrow2as=3u^{2}$ Now, after covering an additional distance s, if velocity becomes v, then, $v^{2}=u^{2}+2a(2s)=u^{2}+4as=u^{2}+6u^{2}=7u^{2}$ $\therefore v=\sqrt{7}u.$