Answered

At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

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.

Sagot :

AL2006

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)


Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.