The displacement is given by $x = 2t^2 + t + 5$. The acceleration at $t = 2 \ s$ is:

If $v = x^2 - 5x + 4$,find the acceleration of the particle when the velocity of the particle is zero.

If velocity and acceleration are in opposite directions in one-dimensional motion,what happens to the magnitude of velocity?

If the displacement ($s$ in metre) of a moving particle in terms of time ($t$ in second) is $s = t^3 - 6t^2 + 18t + 9$,then the minimum velocity attained by the particle is (in $m \ s^{-1}$)

What can be the angle between velocity and acceleration for the motion on a straight line? Explain with an example.

$A$ particle moves along the $x$-axis and its displacement $x$ varies with time $t$ according to the equation $x = c_0(t^2 - 2) + c(t - 2)^2$,where $c_0$ and $c$ are constants of appropriate dimensions. Which of the following statements is correct?