Topic: Engineering Mechanics
A particle starts with a velocity $2~m/sec$ and moves on a straight-line track with retardation 0.1 $m/sec^{2}$. The time at which the particle is 15m from the starting point would be
- A. 10 sec
- B. 20 sec
- C. 50 sec
- D. 40 sec
Answer: A. 10 sec
Explanation: Given initial velocity $u=2~m/s$, retardation $a=-0.1~m/s^{2}$, distance $s=15~m$. Using $s=ut+\frac{1}{2}at^{2}$, we have $15=2t-\frac{1}{2}(0.1)t^{2}$. Simplifying gives $15 = 2t - 0.05t^{2} \implies t^{2} - 40t + 300 = 0$. Factoring, $(t-10)(t-30)=0$. The time is 10 seconds.