A subway train accelerates from rest at one station at a rate of 1.30 m/s^2 for half of the distance to the next station, then decelerates at this same rate for the second half of the distance. If the stations are 3200m apart, find the time of travel (in seconds) between the two stations.

Question

Physics

Blackburn

1 March, 00:49

A subway train accelerates from rest at one station at a rate of 1.30 m/s^2 for half of the distance to the next station, then decelerates at this same rate for the second half of the distance. If the stations are 3200m apart, find the time of travel (in seconds) between the two stations.

+5

Answers (1)

Know the Answer?

Dayami Stark · Answer 1

This problems a perfect application for this acceleration formula:

Distance = (1/2) (acceleration) (time) ².

During the speeding-up half: 1,600 meters = (1/2) (1.3 m/s²) T²

During the slowing-down half: 1,600 meters = (1/2) (1.3 m/s²) T²

Pick either half, and divide each side by 0.65 m/s²:

T² = (1600 m) / (0.65 m/s²)

T = square root of (1600 / 0.65) seconds

Time for the total trip between the stations is double that time.

T = 2 √ (1600/0.65) = 99.2 seconds (rounded)