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

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 $m/s^2$)

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

Match the following columns.

Column $I$	Column $II$
$(A)$ $\frac{dv}{dt}$	$(p)$ Acceleration
$(B)$ $\frac{d|v|}{dt}$	$(q)$ Magnitude of acceleration
$(C)$ $\frac{dr}{dt}$	$(r)$ Velocity
$(D)$ $\left|\frac{dr}{dt}\right|$	$(s)$ Magnitude of velocity

The acceleration-time graph is given. If the initial velocity is $5\,m/s$,then the velocity after $2\,s$ is.......$m/s$.

The relation between the displacement $x$ (in metre) and the time $t$ (in second) of a particle is $t = 2x^2 + 3x$. If the displacement of the particle is $25 \ cm$ from the origin $(x = 0)$, then the acceleration of the particle is: