A police jeep is chasing a culprit going on a motorbike. The motorbike crosses a turning at a speed of 72 km h^-1. The jeep follows it at a speed of 90 km h^-1, crossing the turning 10 s later than the bike. The distance (in km) from turning point at  which the police catch the culprit is ________ .

${\mathrm{v}}_{\mathrm{p}}=90{\mathrm{kmh}}^{-1}=25{\mathrm{ms}}^{-1},{\mathrm{v}}_{\mathrm{c}}=72{\mathrm{kmh}}^{-1}=20{\mathrm{ms}}^{-1}$
In 10 s  culprit reaches point B from A. 

Distance covered by culprit,$\mathrm{s}=\mathrm{vt}=20\times 10=200\mathrm{m}.$
At time t = 10 s, the police jeep is 200 m behind the culprit. Relative velocity between jeep and culprit is $25-20=5{\mathrm{ms}}^{-1}$
Time $=\frac{\mathrm{s}}{\mathrm{v}}=\frac{200}{5}=40\mathrm{s}$
In 40 s, the police jeep will move from A to a distance s where, $\mathrm{s}=\mathrm{vt}=25\times 40=1000\mathrm{m}=1.0\mathrm{km}$ away. The jeep will catch the bike 1 km far from the turning point.