(a) A light-rail commuter train accelerates at a rate of 1.15 m/s^{2}. How long does it take it to reach its top speed of 80.0 km/h starting from rest?

s (b) The same train ordinarily decelerates at a rate of 1.65 m/s^{2}. How long does it take to come to a stop from its top speed? s (c) In emergencies the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in m/s^{2}? m/s^{2}

(a) within a constant acceleration and zero start speed, speed is directly proportional to the time: v = a*t, so t = v/a = 80/1.15 = approx. 69.565 (s).

(b) here v = v0 - a*t. Stop occurs when v=0, so 0 = v0 - a*t, t = v0/a = 80/1.65 = approx. 48.485 (s).

(c) here we have to find a from the same equation 0 = v0 - a*t. So a = v0/t = 80/8.30 = approx. 9.64 (m/s^2).