The correct position $(x)$ - time $(t)$ graph for a particle moving with negative acceleration is

When are the average and instantaneous accelerations equal?

Draw the $x-t$ graphs for positive,negative,and zero acceleration.

$A$ 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?

$A$ 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

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 $m/s^2$)