Honors Physics
posted by Mathew
An automobile accelerates from rest at
1.3 m/s
2
for 19 s. The speed is then held
constant for 23 s, after which there is an acceleration of −0.9 m/s
2
until the automobile
stops.
What total distance was traveled?
Answer in units of km

bobpursley
while accelerating:
distance=1/2 1.3 19^2
finalvelocity=1.3*18
constant speed period.
distance= finalvelocityabove*23 seconds
deaccelerating period.
Vf^2=Vi^2 + 2ad where Vf=0 and Vi equals the finalvelocityabove. Solve for distance.
time deaccelerating: 0=Vi+at solve for t
add the total distances, and if you need, add the total times. 
Mathew
I am still confused. For finalvelocity, where did you get 18?
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