Question

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

• A. 650 m
• B. 750 m
• C. 700 m
• D. 800 m

Answer 1

Answer: (B)

Initial velocity $(u)=0$,
Acceleration $\left(a_1\right)=2m/s^2$ and
time during acceleration $\left(t_1\right)=10$ sec.
Time during constant velocity $\left(t_2\right)=30$ sec
and retardation $\left(a_2\right)=-4m/s^2$
$(-$ ve sign due to retardation).
Distance covered by the particle during acceleration,
$s_1=u{t}_1+\frac{1}{2}{a}_1{t}_1^2=(0\times 10)+\frac{1}{2}\times 2\times (10{)}^2$
$=100m\dots$ (i)
And velocity of the particle at the end of acceleration,
$v=u+a_1{t}_1=0+(2\times 10)=20m/s$.
Therefore distance covered by the particle during constant velocity $\left(s_2\right)$
$=v\times t_2=20\times 30=600m\dots$ (ii)
Relation for the distance covered by the particle during retardation $\left(s_3\right)$ is
$v^2=u_2+2a_2{S}_3$
or, $(0{)}^2=(20{)}^2+2\times (-4)\times s_3=400-8s_3$ or,
$s_3=400/8=50m\dots$ (iii)
Therefore total distance covered by the particle
$s=s_1+s_2+s_1$
$=100+600+50=750m$