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A Thief in a stolen car crashes through the police check post at his top speed of 90 km/h . A motorcycle cop reacting after 2 seconds accelerates from rest at 5 m/s 2 , his top speed being 108 km/h the distance from the checkpost at which the thief is Caught is k × 250 metre then k = ? (Answer lies from 0 to 9 ) #please explain this question without using relative velocity concept##

A Thief in a stolen car crashes through the police check post at his top speed of 90 km/h . A motorcycle cop reacting after 2 seconds accelerates from rest at 5 m/s2 , his top speed being 108 km/h the distance from the checkpost at which the thief is Caught is k × 250 metre then k = ?                   (Answer lies from 0 to 9 ) 
#please explain this question without using relative velocity concept## 

Grade:11

2 Answers

Arun
25750 Points
4 years ago
 

After 2s the thief would be at a distance of 50m fromthe police car.We shall consider this to be t=0.

let after a time t the distance be max.

then the distance between the check post and the thief would be 50+25t.......(i)

the police car would be at a distance of 2.5t^2from the check post...........(ii)

 then the distance between them would be x=50+25t-2.5t^2

this would be max when dx/dt=0

which implies dx/dt=25-5t=0

therefore t=5s.hence x=50+125-62.5=112.5(ans)

Vikas TU
14149 Points
4 years ago
Dear student 
speed(car) = 90 km/h 
=> 25 m/s 
speed(police) max = 108 km/h 
=> 30 m/sec 
time(police) to get to 25 m/s = 25/5 = 5 s 
distance will increase until the police speed reaches 25 m/s: 
d(thief) = s * t_total 
= 25 (5 + 2) =175 m 
d(police) = 1/2 * a * t^2 
= 1/2 * 5 * 5^2 = 62.5 m 

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