If the velocity $v$ of a particle moving along a straight line decreases linearly with its displacement $s$ from $20\,ms ^{-1}$ to a value approaching zero at $s=30\,m$, then acceleration of the particle at $s=15\,m$ is $........$
$\frac{2}{3}\,ms ^{-2}$
$-\frac{2}{3}\,ms ^{-2}$
$\frac{20}{3}\,ms ^{-2}$
$-\frac{20}{3}\,ms ^{-2}$
A car, starting from rest, accelerates at the rate $f$ through a distance $S$, then continues at constant speed for time $t$ and then decelerates at the rate $\frac{f}{2}$ to come to rest. If the total distance traversed is $15S$, then
Stopping distance of vehicles : When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity $(v_0)$ and the braking capacity, or deceleration, $-a$ that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of $v_0 $ and $a$.
The displacement $(x)$ - time $(t)$ graph of a particle is shown in figure. Which of the following is correct?