The train moves along a curvature with a radius of 400 m, and its tangential acceleration is 0.2 m / s2

The train moves along a curvature with a radius of 400 m, and its tangential acceleration is 0.2 m / s2. Determine its normal and full acceleration at the moment when its speed is 10m / s.

Task data: r (radius of curvature) = 400 m; аτ (tangential acceleration) = 0.2 m / s2; V (constant speed of the train at the moment in time) = 10 m / s.

1) We define the centripetal acceleration of the train at the moment in time under consideration: an = V ^ 2 / r = 10 ^ 2/400 = 0.25 m / s2.

2) We calculate the total acceleration of the train: a = √ (an2 + aτ2) = √ (0.25 ^ 2 + 0.2 ^ 2) = 0.32 m / s2.

Answer: The full acceleration of the train at the moment in question will be equal to 0.32 m / s2.