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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
                    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
 
2 months ago
archit
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,
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)t2-8t-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,
b2-4ac
64+4(0.1-a)*40
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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