Question

A particle starts from rest and has an acceleration of 2 $$m/s^2$$for 10 s. After that, it travels for 30s, with constant speed and then undergoes a retardation of 4 $$m/s^2$$ and comes back to rest. The total distance covered by the particle is
A_ 650 m
B- 700 m
C- 750 m
D- 800 m

Answer 1

Hint: The particle starts from rest and so its initial velocity will be zero. Then it moves with a constant acceleration for some time. Then it undergoes constant retardation. Retardation is negative acceleration. We can solve this problem easily by using Newton’s equations of motion.

Complete step by step answer:
For first leg of the journey:
Initial velocity, u= 0
Acceleration, a= 2 $$m/s^2$$
Time, t= 10 s
Using: $$s=ut+{\displaystyle \frac{a{t}^2}{2}}$$
$$\begin{gathered} \Rightarrow s=0+{\displaystyle \frac{2\times {10}^2}{2}} \\ \therefore s=100m \end{gathered}$$
For second leg of the journey:
We need to first find the velocity of the body at the end of 10s,
$$\begin{gathered} v=u+at \\ \Rightarrow v=0+2\times 10 \\ \therefore v=20m/s \end{gathered}$$
Now, the body moves with this velocity for 30s,
$$\begin{gathered} s=vt \\ \Rightarrow s=20\times 30 \\ \therefore s=600m \end{gathered}$$

For third leg of the journey:
Initial velocity, u= 20 m/s
Final velocity, v= 0
Acceleration, a= - $$4m/s^2$$
$$\begin{gathered} v^2-u^2=2as \\ \Rightarrow 0-{20}^2=2\times -4s \\ \Rightarrow -400=-8s \end{gathered}$$
So, the total distance covered will be (100+600+50) m= 750m.

So, the correct answer is “Option C”.

Note:
While doing problems involving Newton’s equations of motion, we have to keep in mind the sign conventions. If a body is accelerating then its acceleration is positive and if the body is coming to stop after some time then it is accelerating and its acceleration is negative. Also, all the units to be used must be in SI units to avoid any mistake. This has to be particularly noted that time can never be negative and if in our answer time is coming in negative then it means we have committed a mistake.