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November 26, 2014

November 26, 2014

Posted by **Mathew** on Tuesday, September 27, 2011 at 10:59pm.

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

- Honors Physics -
**bobpursley**, Tuesday, September 27, 2011 at 11:18pmwhile 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.

- Honors Physics -
**Mathew**, Tuesday, September 27, 2011 at 11:58pmI am still confused. For finalvelocity, where did you get 18?

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