A train that is travelling at 130 km/h applies its brakes as it reaches point A and slows down with a constant deceleration. Its decreased velocity is observed to be 96 km/h as it passes a point 0.8 km beyond A. A car moving at 80 km/h passes point B at the same instant that the train reaches point A. In an unwise effort to beat the train to the crossing, the driver ’steps on the gas’. Calculate the constant acceleration ‘a’ that the car must have in order to beat the train to the crossing by 4 s and find the velocity v of the car as it reaches the crossing. (Kinematics of particles)
SOLUTION:
Since all objects in the problem move with constant acceleration, we may refer to the instant formula applied to the constant acceleration case, namely,
v = v o + at
v 2 = v o 2 + 2a (s – s o )
s = s o + v o t + at 2 /2
For the train, the driver applies the brake causing the constant deceleration. This makes the velocity being reduced from 130 to 96 km/h for the distance traveled 0.8 km.
Therefore the deceleration may be determined:
96^2 = 130^2 + 2a × 0.8
=> a = −4802.5 km/h 2
By this deceleration, the train would reach the intersection at the time counted from the time it reaches point A
1.6 = 130t − 4802.5t^2 /2
=> t = 0.0189 hour or 68.11 s
For the car to beat the train to the crossing by 6 s, the time it must spent counting from point B must be,
t = 68.11 − 6
= 62.11 s or 0.0173 hour
To achieve this, the car must be accelerated by the constant value of
2 = 80 × 0.0173 + a × 0.0173 2 /2
=> a = 4116.40 km/h 2
= 0.317 m/s 2
From this action, the car would have the velocity of
v = 80 + 4116.4 × 0.0173
= 151.21 km/h
= 42.004 m/s, When it reaches the crossing.
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