physics ch3-ch4

Question

A train slows down as it rounds a sharp horizontal turn, going from 96.0 km/h to 52.0 km/h in the 16.0 s that it takes to round the bend. The radius of the curve is 160 m. Compute the acceleration at the moment the train speed reaches 52.0 km/h. Assume the train continues to slow down at this time at the same rate.

magnitude	m/s2
direction	° backward (behind the radial line pointing inward)

Answer 1

First, we need to convert the speeds from km/h to m/s:

Initial speed: v_i = 96.0 km/h = 26.67 m/s Final speed: v_f = 52.0 km/h = 14.44 m/s

The time taken to round the bend is: t = 16.0 s

The radius of the curve is: r = 160 m

We can use the formula for centripetal acceleration to find the acceleration of the train at the moment its speed reaches 52.0 km/h:

a = v^2 / r

where v is the speed of the train and r is the radius of the curve.

At the moment the train speed reaches 52.0 km/h, the speed is:

v = 14.44 m/s

Plugging in the values, we get:

a = v^2 / r = (14.44 m/s)^2 / 160 m = 1.30 m/s^2

Therefore, the acceleration of the train at the moment its speed reaches 52.0 km/h is 1.30 m/s^2.

Since the train continues to slow down at the same rate, we can assume that its acceleration remains constant. The direction of the acceleration is opposite to the direction of the train's motion, which is behind the radial line pointing inward. This is because the acceleration is providing the centripetal force that is necessary to keep the train moving in a circular path.

To find the angle between the acceleration and the radial line pointing inward, we can use the tangent function:

tan(theta) = a / r

where a is the acceleration and r is the radius of the curve.

Plugging in the values, we get:

tan(theta) = 1.30 m/s^2 / 160 m = 0.008125

Taking the inverse tangent, we get:

theta = tan^-1(0.008125) = 0.466 degrees

Therefore, the direction of the acceleration is 0.466 degrees backward (behind the radial line pointing inward).