Subject: General Aptitude · Chapter: Time and Distance · Exam: 2007 · Marks: · Difficulty:
A bus moving at a speed of 24 m/s begins to slow at a rate of 3 m/s each second. How far does it go before stopping?
(a)48 m
(b)60 m
(c)72 m
(d)96 m
Answer
Answer (as printed): D
Explanation
This is a question on uniform retardation (as it is given that the car slows down at a fixed rate) If $v$ is the final velocity, $u$ is the initial velocity, $a$ is the uniform acceleration (or retardation), $t$ is the time and $s$ is the distance covered, we have : $$v=u+a t \quad \text { and } \quad s=u t+\frac{1}{2} a t^{2}$$ Here, $v=0, u=24 \mathrm{m} / \mathrm{s}, a=-3 \mathrm{m} / \mathrm{s}^{2}$ $$\begin{aligned} & \therefore \quad 0=24-3 t \Rightarrow 3 t=24 \Rightarrow t=8 . \\ & \text { And, } s=\left[24 \times 8+\frac{1}{2} \times(-3) \times 8^{2}\right] \mathrm{m}=(192-96) \mathrm{m}=96 \mathrm{m} . \end{aligned}$$