The acceleration of a particle is increasing linearly with time $t$ as $a = 6t$. The particle starts from the origin with an initial velocity $u = 10 \ m/s$. The distance travelled by the particle after $t = 2 \ s$ will be: (in $m$)

Two bodies $A$ and $B$ move in the same straight line starting from the same position. Body $A$ moves with a constant velocity $u$ and body $B$ moves with a constant acceleration $a$ starting from rest. When their velocities become equal,the distance between them is:

$A$ particle is projected with velocity $v_0$ along the $x$-axis. The deceleration of the particle is proportional to the square of the distance from the origin,i.e.,$a = -\alpha x^2$. The distance at which the particle stops is:

$A$ particle starts from the origin at time $t=0$ and moves in the positive $x$-direction. Its velocity $v$ varies with time as $v=10t \text{ cm/s}$. The distance covered by the particle in $8 \text{ s}$ will be: (in $\text{ cm}$)

$A$ 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 $m$)?

If a body starts from rest and travels $120 \,cm$ in the $6^{th}$ second,what is the acceleration in $m/s^2$?