Question

The position of a particle moving along x-axis is given by $x=(t^2-15t+10)\text{m}$, where $t$ is in second. Find the time when particle comes at rest?

• A. $8.5\text{s}$
• B. $7.5\text{s}$
• C. $6.5\text{s}$
• D. $10.5\text{s}$

Answer 1

The correct option is B $7.5\text{s}$

Given, $x=(t^2-15t+10)\text{m}$
For particle at rest, $v=0$
$v=\frac{dx}{dt}=\frac{d}{dt}(t^2-15t+10)$
$v=2t-15$
So, at rest when $v=0\text{m/s}$
$2t-15=0$
$t=7.5\text{s}$