201907251549 Solution to 1984-CE-PHY-II-2

From the graph, the magnitude of velocity $\mathbf{v}$ of an object can be given by a function $v(t)$ of time $t$ :

$v(t)=0.5t+2$ .

The magnitude of displacement $\mathbf{s}$ of the object from time $t=0$ to $t=T$ is a function of time $T$ , given by

$s(T)=\displaystyle{\int_0^T}v(t)\,\mathrm{d}t$ .

In the first 4 seconds ( $T=4$ ),

$\begin{aligned} s(4) & = \int_0^4 (0.5t+2) \,\mathrm{d}t \\ & = \big[0.25t^2+2t\big] \big|_0^4 \\ & = 0.25(4)^2+2(4) \\ & = 12 \end{aligned}$

And the answer is D.