a police man is running toward a thief with a constant acceleration of 0.1 m/s^2. At a particular instant t=0,the policeman is 20 m behind the thief and at the instant the velocity of policeman is 4 m/s. At this instant the thief observed the policeman coming towards him. So he starts running away from the policeman in the same direction with a constant acceleration.Then what is the minimum acceleration with which the thief should run so that he is'nt caught

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Gnaneswar karakavalasa · Answer

For police ,U=4 m/s a=0.1m/s^2 For thief,U=0 m/s a=x (say) Let velocity of police after time t be Vp and that of thief be Vth Vth>Vp or Vth=Vp so that police could not catch the thief Now considering data with respect to thief Up.th=4-0= 4m/s a a.p.th=(0.1-a)m/s^2 Sp.th=20m Vp.th=4m/s^2 Using 3rd equation V^1/2 -U^1/2=2as =》0-16=2 (0.1-a) 20 =》 -16=4-40 a =》 40a=20 =》 a=0.5m/s^2 a p.th=0.1-a

archit · Answer

so in this question , we assume that the police catches the thief at some point at a distance x from the thief and then mathematically force it to given condition. then the distance of police from the point he will catch the thief will be= 20+x; using s=ut+1/2at 2 for police u=4m/s and a=0.1m/s 2 , we get, 20+x=4t+1/2*0.1*t 2 and for thief let the min const acceleration be k ,u=0m/s, x=1/2*k*t, t will be same for both as the time taken to reach the point will be same solving the two we get (0.1-a)t 2 -8t-40=0 we have to force the condition such that they do not meet forcing the condition discriminant we make sure they dont meet as time will have no solution, b 2 -4ac 64+4(0.1-a)*40 we a>=0.5 therefore min acc required to not catch the thief is0.5m/s 2 for police u=4m/s and a=0.1

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