The displacement is given by $x = 2t^2 + t + 5$,the acceleration at $t = 2 \ s$ is ......... $m/s^2$.

$A$ particle has a velocity in the negative direction and a constant acceleration in the positive direction. Match the following columns:

Column $I$	Column $II$
$(A)$ Velocity-time graph	$(p)$ Slope $\rightarrow$ negative
$(B)$ Acceleration-time graph	$(q)$ Slope $\rightarrow$ positive
$(C)$ Displacement-time graph	$(r)$ Slope $\rightarrow$ zero
	$(s)$ $|\text{Slope}| \rightarrow$ increasing
	$(t)$ $|\text{Slope}| \rightarrow$ decreasing
	$(u)$ $|\text{Slope}| \rightarrow$ constant

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

$A$ particle starts from rest. Its acceleration $a$ versus time $t$ is shown in the figure. The maximum speed of the particle will be (in $m/s$)

Motion of a particle is given by the equation $s = (3t^3 + 7t^2 + 3t + 8) \ m$. The acceleration of the particle at $t = 1 \ s$ is: (in $m/s^2$)

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 . . . . . . $\text{m/s}^2$.