Motion in a Lane MCQs for NEET — Physics Questions with Answers

$\frac{dx}{dt}=2at-3b{t}^2\Rightarrow \frac{d^{2}x}{d{t}^2}=2a-6bt=0\Rightarrow t=\frac{a}{3b}$   

The nature of the path is decided by the direction of velocity, and the direction of acceleration. The trajectory can be a straight line, circle or a parabola depending on these factors.

Total distance $=\frac{1}{2}g{t}^2=\frac{25}{2}g$

Distance moved in 3 sec $=\frac{9}{2}g$

Remaining distance $=\frac{16}{2}g$

If t is the time taken by the stone to reach the ground for the remaining distance then

$\Rightarrow \frac{16}{2}g=\frac{1}{2}g{t}^2\Rightarrow t=4\sec$ 

The initial velocity of aeroplane is horizontal, then the vertical component of velocity of packet will be zero.

So $t=\sqrt{\frac{2h}{g}}$  

Horizontal velocity of dropped packet = u

Vertical velocity $=\sqrt{2gh}$

∴ Resultant velocity at earth $=\sqrt{u^2+2gh}$

Given $a=19.6m/s^2=2g$

Resultant velocity of the rocket after 5 sec

$v=2g\times 5=10gm/s$

Height achieved after 5 sec, $h_1=\frac{1}{2}\times 2g\times 25=245m$

On switching off the engine it goes up to height $h_2$ where its velocity becomes zero.

$0={\left(10g\right)}^2-2g{h}_2\Rightarrow h_2=490m$

∴ Total height of rocket $=245+490=735m$ 

Bullet will take $\frac{100}{1000}=0.1$sec to reach target.

During this period vertical distance (downward)

travelled by the bullet $=\frac{1}{2}g{t}^2=\frac{1}{2}\times 10\times {(0.1)}^{2}m=5cm$

So the gun should be aimed 5 cm above the target.

For vertically upward motion, $h_1=v_{0}t-\frac{1}{2}g{t}^2$ and for vertically down ward motion, $h_2=v_{0}t+\frac{1}{2}g{t}^2$

∴ Total distance covered in t sec $h=h_1+h_2=2v_{o}t$. 

$$\begin{gathered} After time t_1: \\ Time to come back to height h after time t_1=t_2-t_1 \\ Time to reach highest point from height h=time to come back to height h from highest point=\frac{t_2-t_1}{2} \\ Time to reach highest point from ground=\frac{t_2-t_1}{2}+t_1=\frac{t_1+t_2}{2} \\ \Rightarrow \frac{t_1+t_2}{2}=\frac{u}{g} \\ \Rightarrow u=\frac{g{\left(t_1+t\right)}_2}{2} \end{gathered}$$

For first projectile, $h_1=ut-\frac{1}{2}g{t}^2$

For second projectile, $h_2=u(t-T)-\frac{1}{2}g{(t-T)}^2$

When both meet i.e. $h_1=h_2$

$ut-\frac{1}{2}g{t}^2=u(t-T)-\frac{1}{2}g{(t-T)}^2$

⇒ $uT+\frac{1}{2}g{T}^2=gtT$

⇒ $t=\frac{u}{g}+\frac{T}{2}$

and $h_1=u\left(\frac{{\displaystyle u}}{{\displaystyle g}}+\frac{{\displaystyle T}}{{\displaystyle 2}}\right)-\frac{1}{2}g{\left(\frac{{\displaystyle u}}{{\displaystyle 2}}+\frac{{\displaystyle T}}{{\displaystyle 2}}\right)}^2$

$=\frac{u^2}{2g}-\frac{g{T}^2}{8}$