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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
pls approve by clicking yes
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
If this reply is helpful then please click YES
Let p=policeman, t =theifVeocity:v(p)= 10km/hr=10.5/18 m/sec=25/9 m/secSimilarly, v(t)=20/9 m/secAs the velocity is uniform, we can use the formula,dist=v.tLet 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.
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