A particle is moving in a straight line under acceleration as shown in figure. The displacement and distance travelled by particle in first 15 seconds will be (at t = 0,v = 0 )

$\begin{array}{l}\mathrm{a}-0=\frac{10}{5}(\mathrm{t}-5)\mathrm{a}=2\mathrm{t}-10\\ {\int }_0^{\mathrm{v}} \mathrm{dv}={\int }_0^{\mathrm{t}} (2\mathrm{t}-10)\mathrm{dt}\mathrm{v}={\mathrm{t}}^2-10\mathrm{t}( \mathrm{Since} \mathrm{v}=0 \mathrm{at} \mathrm{t}=0)\\ {\int }_0^{\mathrm{s}} \mathrm{dx}={\int }_0^{\mathrm{t}} \left({\mathrm{t}}^2-10\mathrm{t}\right)\mathrm{dt}\mathrm{s}=\frac{{\mathrm{t}}^3}{3}-5{\mathrm{t}}^2\mathrm{t}=15\mathrm{s},\mathrm{s}=0\end{array}$
Here, $\mathrm{v}={\mathrm{t}}^2-10\mathrm{t}=0\Rightarrow \mathrm{t}=10\mathrm{s}$
At t =10s, velocity changes sign

Distance in first 15 s: 

$\begin{array}{l}\mathrm{t}=10\mathrm{s},\mathrm{s}\left(10\right)=\frac{500}{3}\mathrm{m}\\ \mathrm{we} \mathrm{know}, \mathrm{t}=15\mathrm{s},\mathrm{s}\left(15\right)=0\\ \text{ Distance }=\left\{\right\{\mathrm{s}\left(10\right)-\mathrm{s}\left(0\right)\}\mid +\{\mathrm{s}\left(15\right)-5\left(10\right)\}\\ =\left|-\frac{500}{3}\right|+\frac{500}{3}=\frac{1000}{3}\mathrm{m}\end{array}$