A thief in a car moving with a uniform speed passes a stationary policeman who is hidhing behind a wall with a motorcycle. After 2.0s delay(reaction time) the policeman accelerates to his maximum speed of 150km/h in 12 sec and catches the theif at 1.5km beyond the wall. The speed of theif is:-
→ 17.50m/s
→ 34.10m/s
→ 51.20m/s
→ 25.20m/s
Answers
Given info : A theif in a car moving with a uniform speed passes a stationary policeman who is hiding behind a wall with motocycle. after 2 sec, the policeman accelerates to his maximum speed of 150 km/h in 12 sec and catches the theif at 1.5 km/h beyond the wall.
To find : the speed of theif is ..
solution : initial velocity of the policeman, u = 0 m/s
final speed of the policeman, v = 150 km/h = 150 × 5/18 m/s
= 750/18 = 41.67 m/s
time taken , t = 12 sec
acceleration, a = (v - u)/t = 41.67/12 = 3.4725 m/s²
so distance covered by policeman, S = ut +1/2 at²
= 0 + 1/2 × 3.4725 × 12² = 250 m
time taken to reach 1.5 km by the policeman :
1.5 km = 250m + 41.67 t
⇒ 1500 - 250 = 1250 = 41.67t
⇒ t = 1250/41.67 = 30 sec
so the total time taken by the policeman = 2s + 12s + 30s = 44s
now the speed of the thief = distance covered by the thief/time taken
= 1500/44 = 34.1 m/s
therefore the speed of thief is 34.10 m/s