The velocity-time graph of robber's car and a chasing police car are shown in the following graph. Police car crosses the robber's car in time   

From graph, velocity of robber's car = $10 {\mathrm{ms}}^{-1}$

Let police car crosses it after t second.

Distance travelled by robber's car = 10 t m

Police car is moving with a constant acceleration of $1 {\mathrm{ms}}^{-2}$ as it attains a velocity of $10 {\mathrm{ms}}^{-1}$ in 10 s after starting from rest. Distance travelled in t second = $\frac{1}{2}·\mathrm{a}·{\mathrm{t}}^2 = \frac{1}{2}{\mathrm{t}}^2$

When the police car crosses the robber's car, distance travelled by the both cars should be same from the starting
point of chase.

$\therefore \frac{1}{2}{t}^2=10t\Rightarrow t=20 \mathrm{s}$