Tangential and Radial Acceleration

[CactuarEnigma]

A train slows down as it rounds a sharp horizontal turn, slowing from 90.0 km/h to 50.0 km/h in the 15.0 s that it takes to round the bend. The radius of the curve is 150 m. Compute the acceleration at the moment the train speed reaches 50.0 km/h. Assume it continues to slow down at this time at the same rate.

So if at = d|v| / dt, what is the function to evaluate? I think calculus fell out of my head over the summer. Thanks for your time.

 

[Doc Al]

Calculus is not needed, if you are familiar with kinematics and centripetal acceleration. Find the tangential and radial components of the acceleration. The tangential component is uniform (figure it out using kinematics); the radial component depends on the speed and the radius.

 

[quicknote]

The tangential and radial acceleration of an object can be calculated using the following equations:

Tangential acceleration (at) = change in tangential velocity (Δv) / change in time (Δt)

Radial acceleration (ar) = (tangential velocity (v)^2) / radius of curve (r)

In the given scenario, the train is slowing down from 90.0 km/h to 50.0 km/h in 15.0 s. This means that the change in tangential velocity (Δv) is 90.0 km/h - 50.0 km/h = 40.0 km/h. The change in time (Δt) is 15.0 s. Therefore, the tangential acceleration (at) at the moment the train speed reaches 50.0 km/h is:

at = Δv / Δt = 40.0 km/h / 15.0 s = 2.67 km/h/s

To calculate the radial acceleration (ar), we need to first convert the tangential velocity (v) from km/h to m/s, since the radius of the curve (r) is given in meters. Converting 50.0 km/h to m/s, we get 50.0 km/h * (1 m / 3.6 km) = 13.89 m/s.

Now, we can plug in the values into the formula for radial acceleration:

ar = (v^2) / r = (13.89 m/s)^2 / 150 m = 1.22 m/s^2

Therefore, at the moment the train speed reaches 50.0 km/h, the total acceleration is the vector sum of the tangential and radial accelerations, which can be calculated using the Pythagorean theorem:

a = √(at^2 + ar^2) = √((2.67 km/h/s)^2 + (1.22 m/s^2)^2) = 2.87 m/s^2

This means that the train is slowing down at a rate of 2.87 m/s^2 at the moment its speed reaches 50.0 km/h. It will continue to slow down at this rate until it comes to a complete stop.