(a) A light-rail commuter train accelerates at a rate of 1.15 m/s2. How long doe

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(a) A light-rail commuter train accelerates at a rate of 1.15 m/s2. 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/s2. 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/s2?
 m/s2
Sep 8th, 2015

Hello!

(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).

Please ask if something is unclear.
Sep 8th, 2015

Sry,but the answer was wrong

Sep 8th, 2015

I was really wrong... 80 is in km/h and I had to make this to m/s units... sorry.

Do you want me to do this now?

Sep 8th, 2015

Yes

Sep 8th, 2015

1 km/h = 1000m/3600s = 5/18 m/s.

(a) 80*(5/18)/1.15 = approx. 19.32 (s)
(b) 80*(5/18)/1.65 = approx. 13.47 (s)
(c) 80*(5/18)/8.30 = approx. 2.68 (m/s^2).

Sep 8th, 2015

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