Asked by robert

A bus accelerates from rest at 1.5m/s^2 for 12s. It then travels for 25s after which it slows to a stop with an acceleration of magnitude 1.5m/s^2. (a) Find the total distance the bus travels And its average velocity of the trip.
Answered by robert
A bus accelerates from rest at 1.5m/s^2 for 12s. It then travels for 25s after which it slows to a stop with an acceleration of magnitude 1.5m/s^2. (a) Find the total distance the bus travel snd its average velocity of the trip.
Answered by robert
666 meters
Answered by Damon
x1 = (1/2)(1.5)(12^2) = 108 m
v1 = 1.5(12) = 18 m/s

x2 = 108 + 18(25) = 558 m
v2 = v1 = 18

x3 = 558 + 18 t - (1/2)1.5 t^2
v3 = 0
= 18 - 1.5 t
so t = 12 seconds to stop (of course)
then
x3 = 558 + 18(12) - (1/2)(1.5)(144)
= 558 + 216 - 108
= 666 meters
Answered by Steve
a devil of a problem!
Answered by aasd
awdadsad
Answered by hello from 2020
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