Question

A train slows down as it rounds a sharp horizontal turn, slowing from 50.9m/s to 23.9m/s in the 23.9s that it takes to round the bend. The radius of the curve is 249.3m. Compute the magnitude of the total acceleration at the moment the train speed reaches 30.7m/s. Assume that it slows down at a constant rate. Give your answer with two decimal places in m/s².

Answer 1

The magnitude of the total acceleration of the train when its speed reaches 30.7m/s, considering both the tangential and centripetal accelerations, is 5.21 m/s².

Step-by-step explanation:

To compute the magnitude of the total acceleration at the moment the train speed reaches 30.7m/s, we must consider both the tangential acceleration due to the train slowing down and the centripetal acceleration due to the curvature of the track.

Firstly, we find the tangential acceleration (at) by using the initial and final speeds (vi and vf respectively) and the time interval (t):

at = (vf - vi) / t

Inserting the given values:

at = (23.9 m/s - 50.9 m/s) / 23.9 s

at = -1.13 m/s2

Next, we find the centripetal acceleration (ac) at the speed of 30.7m/s:

ac = v2/r

ac = (30.7 m/s)2/249.3 m

ac = 3.79 m/s2

Now, we calculate the magnitude of the total acceleration (a) using the Pythagorean theorem, combining tangential and centripetal accelerations:

a = √(at2 + ac2)

a = √((-1.13 m/s2)2 + (3.79 m/s2)2)

a = √(12.84 + 14.36)

a = √(27.20)

a = 5.21 m/s2

Therefore, the magnitude of the total acceleration of the train when its speed is 30.7m/s is 5.21 m/s2.