A thief is spotted by a policeman from a distance of 100 metres. when the policeman starts the chase, the thief also starts running. If the spped of the thief be 8km/hr and that of policeman 10km/hr, how far the thief will have run before he is overtaken?

their time will same so by equating

distanceby theif/speed of theif = distance by police/speed of police

(.1+x)/8 = x/10

by simplifying x = .5 km

hence theif will move .1+.5=.6 km = 600mt

400m .4km

let after time t the policeman will oertake thief and the distance from the current position of thief be x km

then at time when policeman will overtake the thief the time taken by the policeman to run (x+.1)km = xkm covered by thief thus on equating the time expression for both policeman and thief we get the required answer

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Let us assume that the time taken by the policeman to catch the thief is t hrs.

Hence, after time t, the thief will be at a distance of t*8 km from where he started. At the same time he will be at a distance of (t*8) + 0.1 km from where the police started.

As the police catches the thiefr in time t, he will have to run a distance of (t*8) + 0.1 km.

Now, distance run by police in time t= 10*t km.

Therefore, 10t = 8t + 0.1 => 2t= 0.1 => t=0.05 hrs

Distance run by thief in 0.05 hrs = 8*0.05 = 0.4km = 400m

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Let p=policeman,  t =theif
Veocity:v(p)= 10km/hr=10.5/18 m/sec=25/9 m/sec
Similarly, v(t)=20/9 m/sec
As the velocity is uniform, we can use the formula,dist=v.t
Let policeman overtakes the theif at a dist. d from the theif, so 
the time taken to meet at that point will be same for both of them.
i.e. t(p)=t (t)
Therefore, (100+d)/v (p)=d/v(t)         ( time= dist/v  and policeman has to cover 100m more than theif)
 by putting the values  of velocities and solving we get d=400 m.