A train slows down as it rounds a sharp horizontal turn, going from 86.0 km/h to 56.0 km/h in the 18.0 s that it takes to round the bend. The radius of the curve is 160 m. Compute the acceleration at the moment the train speed reaches 56.0 km/h. Assume the train continues to slow down at this time at the same rate.

The acceleration of the train is 1.581 m/s² inward.

Explanation:

Given;

initial velocity of the train, u = 86.0 km/h = 23.889 m/s

final velocity of the train, v = 56.0 km/h = 15.556 m/s

change in time, Δt = 18 s

The total acceleration of particles moving along a curved path is given as vector sum of the tangential acceleration and radial acceleration

where;

is the tangential acceleration

is radial acceleration

Therefore, the acceleration at the moment the train speed reaches 56.0 km/h is 1.581 m/s² inward.

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