Robbers in a car traveling at 20m/s pass a police man at rest . The police man immediately start chasing the robbers . The police man accelerates at 3m/s^2 for 12 second and thereafter travels at a constant speed. Calculate the distance covered by the police man before he overtakes the car.
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Here, u=0,t=12 sec , a=3 m/s²
Distance covered by the policeman in 12 seconds is:-
We know that,
s=ut + 1/2at²
→0×12 +1/2(3)(12)²
→1/2(3)(144)
→3×72
→216 m
v = u +at
→0 + 3(12)
→36 m/s
The distance covered by the robber in 12 seconds is :-
Distance = speed × time
→20 m/s × 12 s
→240 m
Suppose the policeman overtakes in 12+t seconds
Then distance = 216 + 36t
Then distance covered by the robber in the same time is
→Speed × time = 20 ( 12 + t )
216 + 36 t = 240 + 20 t
36t-20t = 240-216
16t = 24
t = 24/16
t = 3/2 = 1.5 seconds
The distance covered by the policeman before he overtakes the car in 12 + 1.5 = 13.5 seconds is
216+36t
→216 + 36(1.5)
→216 + 54
→270 m
Distance covered by the policeman in 12 seconds is:-
We know that,
s=ut + 1/2at²
→0×12 +1/2(3)(12)²
→1/2(3)(144)
→3×72
→216 m
v = u +at
→0 + 3(12)
→36 m/s
The distance covered by the robber in 12 seconds is :-
Distance = speed × time
→20 m/s × 12 s
→240 m
Suppose the policeman overtakes in 12+t seconds
Then distance = 216 + 36t
Then distance covered by the robber in the same time is
→Speed × time = 20 ( 12 + t )
216 + 36 t = 240 + 20 t
36t-20t = 240-216
16t = 24
t = 24/16
t = 3/2 = 1.5 seconds
The distance covered by the policeman before he overtakes the car in 12 + 1.5 = 13.5 seconds is
216+36t
→216 + 36(1.5)
→216 + 54
→270 m
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