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
Concept. Distances in successive equal seconds differ by $a \times 1^2$; then use $s = \dfrac{u^2}{2a}$ for the total stopping distance.
Step 1. Retardation. $a = 22 - 18 = 4\ \text{m/s}^2$.
Step 2. Speed at the start. $22 = u - \dfrac12(4)(1) \Rightarrow u = 24$ m/s.
Step 3. Total stopping distance. $\dfrac{24^2}{2 \times 4} = 72$ m.
Step 4. Further distance. $72 - (22 + 18) = 32$ m.
Answer. $32$ m (Option A).
Solution by Sreeraj P, M.Sc Physics