Question

A truck starts from rest and accelerates uniformly at $2{m{s^{-2}}}$. At t=10s, a stone is dropped by a person standing on the top of the truck (6m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t=11s ? (Neglect air resistance.)

Answer 1

This is a problem on relative motion. The stone thrown in the 10th second follows a parabolic path during fall, as it has both horizontal and the vertical velocity. The horizontal component is the result of the truck and the vertical component is the result of gravity.

Formulas used:
First law of motion: $v=u+at$where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time.
Resultant velocity $v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}$ where ${{v}_{x}}$ is the horizontal component and ${{v}_{y}}$ is the vertical component.

Complete step by step answer:
We know that the initial velocity of the truck u=0
Acceleration $a=2m/{{s}^{2}}$
And time t=10s
From the first equation of motion,
$\begin{aligned} & v=u+at \\\ & \Rightarrow 0+2\times 10=20m/s \\\ \end{aligned}$
The final velocity of the truck and the stone is $20m/s$
At t=11s the horizontal component ${{v}_{x}}$remains unchanged in the absence of air resistance,
Hence, ${{v}_{x}}=20m/s$
The vertical component can be calculated by
$\begin{aligned} & {{v}_{y}}=u+{{a}_{y}}t \\\ & \Rightarrow t=11-10=1s \\\ & \Rightarrow {{a}_{y}}=g=10m/{{s}^{2}} \\\ & \Rightarrow {{v}_{y}}=0+10\times 1=10m/s \\\ \end{aligned}$
Therefore resultant velocity is given by
$\begin{aligned} & v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}} \\\ & \Rightarrow v=\sqrt{{{20}^{2}}+{{10}^{2}}}=22.36m/s \\\ \end{aligned}$
If the angle made by v with ${{v}_{x}}$be $\theta$
Hence,
$\tan \theta =\dfrac{{{v}_{y}}}{{{v}_{x}}} \\\ \Rightarrow \theta ={{\tan }^{-1}}(\dfrac{{{v}_{y}}}{{{v}_{x}}}) ={{\tan}^{-1}}(\dfrac{10}{20})=26.570 \\\$

Also, whenever the stone is dropped from the truck, the horizontal force is zero but the force of gravity acts on it. Hence its acceleration is $10m/{{s}^{2}}$ acting vertically downward.

Note: The idea of relative motion in two dimensions is quite alike to that of the relative velocity in a straight line. According to questions, we take more than one moving object in a frame which might be stationary with respect to another point.