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Grade: 9


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

2 months ago

Answers : (2)

Gnaneswar karakavalasa
13 Points
For police ,U=4 m/s
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 4m/s
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
2 months ago
13 Points
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+1/2at2
for 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,
t will be same for both as the time taken to reach the point will be same
solving the two we get
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,
we a>=0.5
therefore min acc required to not catch the thief  is0.5m/s2
for police u=4m/s and a=0.1
2 months ago
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