Question

Freight trains can produce only relatively small accelerations. (a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s² for 8.00 min, starting with an initial velocity of 4.00 m/s? (b) If the train can slow down at a rate of 0.550 m/s², how long will it take to come to a stop from this velocity? (c) How far will it travel in each case?

a) (a) 240 m/s; (b) 8.2 s; (c) 8160 m
b) (a) 0.70 m/s; (b) 7.3 s; (c) 20.2 m
c) (a) 24 m/s; (b) 8.5 s; (c) 1020 m
d) (a) 0.30 m/s; (b) 14.5 s; (c) 85 m

Answer 1

The final velocity of the freight train after accelerating for 8.00 minutes at 0.0500 m/s² is 28 m/s. It will take approximately 50.91 seconds to come to a stop from this velocity, decelerating at 0.550 m/s². The train will travel 7680 m while accelerating and approximately 706.96 m while decelerating.

Step-by-step explanation:

To answer the questions about the freight train's acceleration, deceleration, and distances traveled, we need to use the kinematic equations of motion. These allow us to calculate the final velocity, time taken to stop, and total distance covered.

Part (a)

Final velocity (v) can be found using the formula: v = u + at, where 'u' is the initial velocity, 'a' is the acceleration, and 't' is the time.

First, convert 8.00 minutes to seconds:
8.00 min × 60 s/min = 480 s

Then, calculate the final velocity:
v = 4.00 m/s + (0.0500 m/s² × 480 s) = 4.00 m/s + 24 m/s = 28 m/s.

Part (b)

Time to come to a stop (t) can be found using: t = v/a.

t = 28 m/s / 0.550 m/s² = 50.91 seconds (approximately).

Part (c)

Distance traveled during acceleration (d) can be found using: d = ut + 0.5at².

d = (4.00 m/s × 480 s) + 0.5 × 0.0500 m/s² × (480 s)² = 1920 m + 5760 m = 7680 m.

Distance traveled during deceleration (d) can be found using: d = ut - 0.5at².

d = 28 m/s × 50.91 s - 0.5 × 0.550 m/s² × (50.91 s)² = 1425.48 m - 718.52 m = 706.96 m (approximately).