A train slows down as it rounds a sharp horizontal curve,...

A train slows down as it rounds a sharp horizontal curve, slowing from $90.0 \mathrm{~km} / \mathrm{h}$ to $50.0 \mathrm{~km} / \mathrm{h}$ in the $15.0 \mathrm{~s}$ that it takes to round the curve. The radius of the curve is $150 \mathrm{~m}$. Compute the acceleration at the moment the train speed reaches $50.0 \mathrm{~km} / \mathrm{h}$. Assume that the train slows down at a uniform rate during the $15.0-\mathrm{s}$ interval.

Key Concepts

Kinematics of Uniformly Accelerated Motion

This concept involves the equations of motion that describe the behavior of an object under constant acceleration. In the context of the problem, it is used to determine changes in speed over time, which then feed into calculations for other aspects of the motion, like tangential acceleration.

Vector Decomposition of Acceleration

When an object moves in a curved path with changing speed, its overall acceleration can be decomposed into two perpendicular components: the tangential acceleration (affecting speed) and the centripetal (or radial) acceleration (affecting direction). This decomposition is crucial in analyzing motion along a curved path since each component can be treated independently.

Centripetal Acceleration

In circular motion, an object moving along a curved path experiences an inward acceleration known as centripetal acceleration. This acceleration is responsible for changing the direction of the velocity vector so that the object continues to follow a curved trajectory, and its magnitude is determined by the square of the speed divided by the radius of the curve.

Uniform Tangential Acceleration

Uniform tangential acceleration refers to a constant rate of change in the object's speed along the tangent to the path of motion. In problems involving deceleration or acceleration along a curved path, this component affects the magnitude of the velocity while the centripetal component affects its direction.

Transcript

0:00 Hello.

00:02 In this question, we have a train that is moving on a rail until it faced the horizontal curve, and it starts to slow down in this curve from 90 kilometers per hour until it reached 50 kilometers per hour at the end of the curve.

00:25 The time that it takes the train to reach from this velocity to this velocity was 15 seconds.

00:31 And also we know that the diameter of this curve or the radius is 150 meters.

00:41 So we're also to calculate the acceleration at the point when the train was moving at 50 kilometers per hour.

00:50 So first of all, we know that in this case we have two accelerations, one that is radial and points towards the center of the curve and the other acceleration will be pattern to the train's velocity or in the theta direction if we use this notation.

01:08 So let's first determine the acceleration that or the tangential acceleration, which is responsible for actually decreasing the speed rather than changing the velocity.

01:22 So here we need to calculate a tangential and basically this is delta v by delta t.

01:31 Assuming that the acceleration was constant throughout the entire 15 seconds.

01:37 So here, the final velocity was 50, and the initial velocity was 90, and let's convert this to be in meters per second instead of kilometers per hours.

01:51 So we need to multiply by 1 ,000 to convert to meters, and divide by 3 ,600 to convert to seconds.

02:03 And this happened over 15 seconds.

02:06 And plugging on these numbers on a catheter, we will find that the answer is negative 0 .74 meters per second.

02:17 And that's in the direction of theta.

02:21 And of course, the negative sign means that it's opposing the motion of the train.

02:25 So here, if we assume that the motion of the train is in this direction, so now we know that the acceleration or the speed, the tangential acceleration factor is pointing in that way...

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