Q79 / 160 · WBJEE 2015 · Mathematics
A particle starts moving from rest from a fixed point in a fixed direction. The distance s from the fixed point at a timet is given by $s=t^{2}+a t-b+17,$ where $a$ and $b$ are real numbers. If the particle comes to rest after 5s at a distance of $s=25$ units from the fixed point, then values of $a$ and $b$ are. respectively
- A10, -33
- B-10,-30
- C-8.33 .
- Correct answerD-10.33
Solution
Given, $s=t^{2}+a t-b+17$
After 5 s. $\left[\frac{d s}{d t}\right]_{t=5}=[2 t+a]_{t=5}=0$
$\Rightarrow \quad a=-10$
At $t=5 \mathrm{s}, s=25$ units
$\therefore \quad 25=5^{2}+5 a-b+17$
$$
\mid \because s=25 \text { and } t=5 \mid
$$
$\Rightarrow \quad 25=25+5(-10)-b+17$
$\Rightarrow \quad 50-17=-b$
$\Rightarrow \quad b=-33$
Hence, the correct answer is option: 'D'.