# At the moment the train's speed is 59km/h, what is the magnitude of the total acceleration?A train slows down at a constant rate as it rounds a sharp circular horizontal turn. It takes 16.3 s to...

At the moment the train's speed is 59km/h, what is the magnitude of the total acceleration?

A train slows down at a constant rate as it rounds a sharp circular horizontal turn. It takes 16.3 s to slow down from 74 km/h to 40 km/h.  The radius of the curve is 158 m.

justaguide | College Teacher | (Level 2) Distinguished Educator

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The train slows down at a constant rate as it rounds a sharp circular horizontal turn. It takes 16.3 s to slow down from 74 km/h to 40 km/h.  The radius of the curve is 158 m.  At the moment the train's speed is 59 km/h, what is the magnitude of the total acceleration?

The train is not moving in uniform circular motion. Due to this there is a tangential acceleration acting on it as well as a radial acceleration directed towards the center of the circular path it is moving in.

Converting the velocity of the train to m/s, 74 km/h = 20.56 m/s and 40 km/h = 11.11 m/s. The tangential acceleration is (11.11 - 20.56)/16.3 = -0.579 m/s^2. When the velocity of the train is 59 km/h = 16.38 m/s as it is moving in a circular path with radius 158 m, the centripetal acceleration is (16.38)^2/158 = 1.69 m/s^2. The sum of these accelerations that are perpendicular to each other is equal to sqrt(1.69^2 + 0.579^2) = 1.79 m^2/s

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