Q 11-02-041NEETJEE MainEasy
An elevator starts from rest and its speed increases uniformly to $4$ m/s in $3$ s. It moves at this speed until $t = 9$ s and then slows down uniformly to stop at $t = 12$ s. The height through which it rises is
Answer: (B) $36$ m
Concept. Height risen $=$ area under the $v$–$t$ graph (a trapezium).
Step 1. Area.
$$\frac12(3)(4) + (9 - 3)(4) + \frac12(3)(4) = 6 + 24 + 6 = 36\ \text{m}$$
Answer. $36$ m (Option B).
Solution by Sreeraj P, M.Sc Physics