Kinematics-1D

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, $v^{2}=u^{2}+2a(2s)=u^{2}+4as=u^{2}+6u^{2}=7u^{2}$ $\therefore v=\sqrt{7}u.$