Q 11-02-024NEETJEE MainHard
A particle moving in a straight line with uniform retardation covers $22$ m and $18$ m in two successive seconds. How much further will it travel before coming to rest?
Answer: (A) $32$ m
In successive equal seconds, the distances differ by $a \times 1^2$, so the retardation is $a = 22 - 18 = 4\ \text{m/s}^2$.
Let the speed at the start of the first of these seconds be $u$. Distance in that second:
$$22 = u \times 1 - \frac{1}{2}(4)(1)^2 \;\Rightarrow\; u = 24\ \text{m/s}$$
Total distance from that point until rest: $\dfrac{u^2}{2a} = \dfrac{576}{8} = 72$ m.
It has already covered $22 + 18 = 40$ m, so it travels a further $72 - 40 = 32$ m.
Solution by Sreeraj P, M.Sc Physics