In some appropriate units,the time $(t)$ and position $(x)$ relation of a moving particle is given by $t = x^2 + x$. The acceleration of the particle is

In a car race on a straight road,car $A$ takes a time $t$ less than car $B$ at the finish and passes the finishing point with a speed $v$ more than that of car $B$. Both the cars start from rest and travel with constant acceleration $a_1$ and $a_2$ respectively. Then $v$ is equal to

$A$ driver applies the brakes on seeing the red traffic signal $400 \ m$ ahead. At the time of applying brakes,the vehicle was moving with $15 \ m/s$ and retarding at $0.3 \ m/s^2$. The distance of the vehicle from the traffic light one minute after the application of brakes is: (in $m$)

$A$ car is moving with velocity $V$. It stops after applying brakes at a distance of $20 \ m$. If the velocity of the car is doubled,then what distance will it cover after applying the brakes (in $m$)?

$A$ particle with initial velocity $v_0$ moves with constant acceleration $a$ in a straight line. Find the distance travelled in the $n^{th}$ second.

The displacement of a particle is proportional to the cube of time elapsed. How does the acceleration of the particle depend on time?