Question

A particle starts from rest and has an acceleration of $2\, m / s ^{2}$ for $10 \,sec$. After that, it travels for $30\, sec$ 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. 750 m
• C. 700 m
• D. 800 m

Answer 1

The Correct Option is B

Initial velocity $(u)=0$, Acceleration $\left( a _{1}\right)=2 m / s ^{2}$ and time during acceleration $\left( t _{1}\right)= 10$ sec. Time during constant velocity $\left( t _{2}\right)=30$ sec and retardation $\left( a _{2}\right)=-4 m / s ^{2}$ $(-$ ve sign due to retardation). Distance covered by the particle during acceleration, $s_{1}=u t_{1}+\frac{1}{2} a_{1} t_{1}^{2}=(0 \times 10)+\frac{1}{2} \times 2 \times(10)^{2}$ $=100 \,m \ldots$ (i) And velocity of the particle at the end of acceleration, $v=u+a_{1} t_{1}=0+(2 \times 10)=20\, m / s$. Therefore distance covered by the particle during constant velocity $\left(s_{2}\right)$ $=v \times t_{2}=20 \times 30=600\, m \ldots$ (ii) Relation for the distance covered by the particle during retardation $\left( s _{3}\right)$ is $v ^{2}= u _{2}+2 a _{2} S _{3}$ or, $(0)^{2}=(20)^{2}+2 \times(-4) \times s _{3}=400-8 s _{3}$ or, $s _{3}=400 / 8=50 \,m \ldots$ (iii) Therefore total distance covered by the particle $s = s _{1}+ s _{2}+ s _{1}$ $=100+600+50=750 \,m$