The relation between time ' $t$ ' and distance ' $x$ ' is $t=$ $\alpha x^2+\beta x$, where $\alpha$ and $\beta$ are constants. The relation between acceleration $(a)$ and velocity $(v)$ is:
$a=-2 \alpha v^3$
$a=-5 \alpha v^5$
$a=-3 \alpha v^2$
$a=-4 \alpha v^4$
The position$(x)$ of a particle at any time$(t)$ is given by $x(t) = 4t^3 -3t^2 + 2$ The acceleration and velocity of the particle at any time $t = 2\, sec$ are respectively
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 $........$
The acceleration of a train between two stations is shown in the figure. The maximum speed of the train is $............\,m/s$
Consider the acceleration, velocity and displacement of a tennis ball as it falls to the ground and bounces back. Directions of which of these changes in the process