$A$ body is moving with a uniform acceleration covers $40\,m$ in the first $4\,s$ and $120\,m$ in next $4\,s.$ Its initial velocity and acceleration are

$A$ car accelerates from rest at a constant rate $\alpha$ for some time,after which it decelerates at a constant rate $\beta$ and comes to rest. If the total time elapsed is $t$,then the maximum velocity acquired by the car is

$A$ particle of mass $m$ is constrained to move on the $x$-axis. $A$ force $F$ acts on the particle. $F$ always points toward the position labeled $E$. For example,when the particle is to the left of $E$,$F$ points to the right. The magnitude of $F$ is constant except at point $E$ where it is zero. The system is horizontal. $F$ is the net force acting on the particle. The particle is displaced a distance $A$ towards the left from the equilibrium position $E$ and released from rest at $t=0$. Find the minimum time it will take to reach from $x=-A/2$ to $x=0$.

$A$ particle moving along the $X$-axis has acceleration $f$ at time $t$ given by $f=f_0\left(1-\frac{t}{T}\right)$,where $f_0$ and $T$ are constants. The particle at $t=0$ has zero velocity. In the time interval between $t=0$ and the instant when $f=0$,the particle's velocity is

$A$ vehicle moving with $15 \,km/hr$ comes to rest by covering $5 \,m$ distance by applying brakes. If the same vehicle moves at $45 \,km/hr$, then by applying brakes, it will come to rest by covering a distance of: (in $\,m$)

The position of a particle moving along the $x$-axis is given by $x = (-2t^3 + 3t^2 + 5) \ m$. The acceleration of the particle at the instant its velocity becomes zero is ....... $m/s^2$.