A train moving at 30 m s⁻¹ begins to decelerate uniformly at 1.5 m s⁻². At the same instant, a car 150 m behind the train moves in the same direction at constant speed 40 m s⁻¹. The car catches the train after:
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Question

A train moving at 30 m s⁻¹ begins to decelerate uniformly at 1.5 m s⁻². At the same instant, a car 150 m behind the train moves in the same direction at constant speed 40 m s⁻¹. The car catches the train after:

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G1 Physics · Accepted Answer

Take the car's initial position as x = 0. Then x_car = 40t. Train position is x_train = 150 + 30t − 0.75t². Equating gives 0.75t² + 10t − 150 = 0. Multiplying by 4: 3t² + 40t − 600 = 0. The positive root is t = 10 s.