Thursday

March 5, 2015

March 5, 2015

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?

**Answer this Question**

**Related Questions**

Physics - An automobile accelerates from rest at 1.3 m/s 2 for 19 s. The speed ...

physics - An automobile accelerates from rest at 1.3m/s^2 for 19s. The speed is ...

Physics - An automobile accelerates from rest at 2.4 m/s2 for 29 s. The speed is...

physics - An automobile and train move together along parallel paths at 28.8 m/s...

physics - An automobile and train move together along parallel paths at 20.3 m/s...

Physics - An automobile and train move together along parallel paths at 21.8 m/s...

physics - An automobile and train move together along parallel paths at 39.1 m/...

Physics - An automobile and train move together along parallel paths at 31.6 m/s...

Physics - Could you please check my answers and units. Thanxs! 1. An automobile ...

Physics - A motor of a certain elevator gives it a constant upward acceleration ...