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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 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
For police ,U=4 m/s a=0.1m/s^2For 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/sa a.p.th=(0.1-a)m/s^2 Sp.th=20m Vp.th=4m/s^2Using 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
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+1/2at2for police u=4m/s and a=0.1m/s2 ,we get, 20+x=4t+1/2*0.1*t2 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 samesolving the two we get(0.1-a)t2-8t-40=0we have to force the condition such that they do not meet forcing the condition discriminantwe make sure they dont meet as time will have no solution,b2-4ac64+4(0.1-a)*40we a>=0.5therefore min acc required to not catch the thief is0.5m/s2 for police u=4m/s and a=0.1
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