Consider that a truck is moving initially with $54 \,km/h$. It is stopped by the driver after looking at an obstacle with a deceleration of $10 \,m/s^2$. The distance travelled by the truck before coming to rest is (in $\,m$)

$A$ particle moves along a straight line $OX$. At a time $t$ (in seconds),the distance $x$ (in metres) of the particle from $O$ is given by $x = 40 + 12t - t^3$. How long would the particle travel before coming to rest?

$A$ body starts from the origin and moves along the $X$-axis such that the velocity at any instant is given by $v = (4t^3 - 2t)$,where $t$ is in $s$ and velocity is in $m/s$. What is the acceleration of the particle when it is $2\, m$ from the origin? (in $m/s^2$)

The initial velocity of a particle is $u$ (at $t = 0$) and the acceleration is given by $a(t) = at$. Which of the following relations is valid?

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 $a$.

$A$ car starts at time $t=0$ from an initial speed of $10 \,m/s$ and accelerates uniformly with $2 \,m/s^2$ on a straight road for time $0 \leq t \leq 10 \,s$. Let $S_1$ and $S_2$ be the distance covered by the car in time $3 \leq t \leq 4 \,s$ and $4 \leq t \leq 5 \,s$ respectively. The ratio $\frac{S_2}{S_1}$ is