# Alias robert

a car starting from rest is accelerated at 2m/s(square) for 10s, then moves at constant velocity for 20s and then decelerates at 1m/s(square), finally stops. how far does it travel during its trip? please help me! I need in solution for that problem

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1. a = 2
v = 2t + c, where t is in seconds
when t=0, v = 0 , (it started from rest)
0 = 0 + c, c = 0

v = 2t

s = t^2 + k
when t = 0, s = 0 , (hasn't gone anywhere)

after 10 seconds
v = 20 m/s
s = 100 m

now goes for 20 s, at 20 m/s, covering 400 m
so it can covered 500 m, and the speed is 20 m/s

now repeat the process: starting again with a time of t=0, v = 20, s=0 (remembering we have already gone 500 m)
a = -1
v = -t + p
when t = 0, v = 20
20 = 0 + p
so v = -t + 20
s = (-1/2)t^2 + 20t + q
q = 0
s = (-1/2)t^2 + 20t

when it came to a stop, v = 0
0 = -t + 20
t = 20, so it took 20 seconds to stop
in those 20 seconds, the car went another
(-1/2)(20)^2 + 20(20) m
= 200 m

so the whole "trip" was 700 m

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posted by Reiny

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