Question

Velocity in $m/s$ of a particle moving along $x-$axis varies with time as $v=(10+5t-t^2)$. At time $t=0,x=0.$ Acceleration of particle at $t=2s$ is $xm/s^2$, then value of $x$ is

Answer 1

We have $a=\frac{dv}{dt}=5-2t$
At $t=2s,a=5-2\times 2=1m/s^2$.
Thus $x=1.$