posted by Mathew .
An automobile accelerates from rest at
for 19 s. The speed is then held
constant for 23 s, after which there is an acceleration of −0.9 m/s
until the automobile
What total distance was traveled?
Answer in units of km
distance=1/2 1.3 19^2
constant speed period.
distance= finalvelocityabove*23 seconds
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.
I am still confused. For finalvelocity, where did you get 18?