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

Each of the three graphs represents acceleration versus time for an object that already has a positive velocity at time $t_1$. Which graphs show an object whose speed is increasing for the entire time interval between $t_1$ and $t_2$?

The velocity of a particle is given as $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k} \text{ m/s}$. The magnitude of acceleration at point $(1, 2, 4)$ is . . . . . . $\text{m/s}^2$.

The displacement of a particle,moving in a straight line,is given by $s = 2t^2 + 2t + 4$,where $s$ is in metres and $t$ is in seconds. The acceleration of the particle is ........ $m/s^2$.

The velocity $(v)-$ time $(t)$ plot of the motion of a body is shown below:
The acceleration $(a)-$ time $(t)$ graph that best suits this motion is:

The velocity of a particle is given by the equation $v(x) = 3x^2 - 4x$,where $x$ is the distance covered by the particle. The expression for its acceleration is