Catch the thief!

As the policeman is 200 meters behind the thief and he must catch him in 10 seconds , - Distance to be covered by the policeman = 200 + Distance covered by the thief in the given time( s1_ As the velocity of thief is constant , distance covered by him will be equal to velocity $\times$ time = 5 m/s $\times$ 10 s= 50 meters= $s_{1}$ Let the initial velocity of the policeman be "u" ,the time taken be " t" , the acceleration of the policeman be "a". Using the 2nd equation of motion , we get 200+ $s_{1}$ = ut + $\frac{1}{2}$ a $t^{2}$ .

As the policeman is initially at rest, $u=0$ , and as $s{1} = 50 m$ . And as the policeman must catch the thief in 10 seconds, $t = 10$ .

$250=\frac{1}{2} \times a \times 10^{2} \\ 250 = \frac{1}{2}\times a \times 100 \\ 250 = 50a \\ \boxed{a = 5 m/s^2}.$