Car A, moving at a constant speed of 30 m/s, passes a stationary car, B. Immediately after this, car B starts from rest and accelerates uniformly at 2 m/s2 until it reaches a speed of 40 m/s. It then continues forward, maintaining this speed. Find the time taken by car B to overtake car A and the distance from the starting point when the overtaking occurs.
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A car traveling at a constant speed of 30m/s passes a highway patrol car, which is at rest. The police officer accelerates at a constant rate of 3m/s
2
and maintains this rate of acceleration until he pulls next to the speeding car. Assume that the police car starts to move at the moment the speeder passes the police car. What is the time required for the police officer to catch the speeder?
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Solution
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Correct option is
A
20s
The origin of co ordinate system (x-axis), we assume it is the point where speeding car passes the police car
x
speeding
(t)=30t→1
The motion of the speeding car is the motion with constant speed of 30㎧
The motion of the police car is the motion with constant acceleration
∴x
police
(t)=v
initial
(t)+a
2
t
2
v(t)=v
initial
+at
v
2
(t)−v
initial
2
=2ax
police
(t)
a=3㎨
Initial velocity=0=v
initial
∴x
police
(t)=1.5t
2
→2
v(t)=3t
v
2
(t)=6x
police
(t)
x
speeding
(t
0
)=x
police
(t
0
)→3
Substituting equation 1,2
30t
0
=1.5t
0
2
t
0
=
1.5
30
=20s
Thus after 20s the police car catches the speeding car.