Ahigh-speed bullet train accelerates and decelerates at the rate of 4 ft/s^2. its maximum cruising speed is 90 mi/h. suppose that the train starts from rest and must come to a complete stop n 15 minutes. what is the maximum distance it can travel under these conditions?

810 000 ft

Explanation:

First we convert 90 mi/h to ft/s

90 mi/h = 90 mi/h * 5280 ft/1 mi * 1 hr/3600 s = 132 ft/s

To find the distance covered for the train to travel from rest to 132 ft/s we use an equation of linear motion based on the values we have and the measurement we are supposed to find:

x = 0.5at²    [the term for initial velocity is omitted because the train starts

                   form rest]

x = 0.5 * 4 ft/s² * (450 s)² = 405 000 ft    

The maximum distance will be twice this because the train must reach the maximum velocity and again decelerate to zero.

distance covered = 2 * 405 000 ft = 810 000 ft =

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray