# What Is the Acceleration of a Train Slowing Down on a Curve?

• Sunnie
In summary, the train slows down from 84.2 km/h to 50.7 km/h in 15 seconds as it rounds a sharp horizontal turn with a radius of 120 m. The acceleration at the moment the train reaches 50.7 km/h is calculated to be -5.204 m/s2. The direction of the acceleration is 180° backward (behind the radial line pointing inward).
Sunnie
A train slows down as it rounds a sharp horizontal turn, slowing from 84.2 km/h to 50.7 km/h in the 15 s that it takes to round the bend. The radius of the curve is 120 m. Compute the acceleration at the moment the train speed reaches 50.7 km/h. Assume that it continues to slow down at this time at the same rate.

What is the magnitude in m/s2?

What is the direction in __ ° backward (behind the radial line pointing inward)

The formulas that I have is:

Vf= Vi + AT

Rf= Ri + Vi*t + (1/2)A*T^2

A = (Vf-Vi)/t First, solve for A: A = (50.7 km/h - 84.2 km/h) / (15 s) A = -5.204 m/s2 The magnitude of the acceleration is 5.204 m/s2. The direction of the acceleration is 180 ° backward (behind the radial line pointing inward).

Hi there,

I would be happy to help you with this problem! It seems like you have a good understanding of the relevant formulas, but let's go through the problem step by step to make sure we get the correct answer.

First, let's convert the initial and final speeds from km/h to m/s, since the formulas we have use meters per second (m/s).

Initial speed (Vi) = 84.2 km/h = 84.2 * (1000 m / 1 km) * (1 h / 3600 s) = 23.39 m/s
Final speed (Vf) = 50.7 km/h = 50.7 * (1000 m / 1 km) * (1 h / 3600 s) = 14.08 m/s

Next, we can use the formula Vf = Vi + AT to solve for the acceleration (A). We know that Vf = 14.08 m/s, Vi = 23.39 m/s, and the time (T) it takes to round the bend is 15 seconds. So we have:

14.08 m/s = 23.39 m/s + A * 15 s

Rearranging the equation to solve for A, we get:

A = (14.08 m/s - 23.39 m/s) / 15 s = -0.623 m/s^2

So the acceleration at the moment the train speed reaches 50.7 km/h is -0.623 m/s^2.

To find the magnitude of the acceleration, we can take the absolute value of -0.623 m/s^2, which is 0.623 m/s^2.

Now, let's find the direction of the acceleration. The acceleration is negative, which means it is in the opposite direction of the initial velocity. Since the train is moving in a horizontal direction, the acceleration is in the vertical direction, pointing downward.

To find the angle of the acceleration, we need to use some trigonometry. Let's draw a diagram to better visualize the situation:

/|
/ |
/ |
/ |
/____|

From the diagram, we can see that the angle we want to find is the angle between the radial line pointing inward (the radius of the curve) and the direction of the acceleration.

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