Posted by **sam** on Friday, September 14, 2012 at 11:39pm.

An automobile accelerates from rest at 1.3m/s^2 for 19s. The speed is then held constant for 23s, after which there is an acceleration of -0.9m/s^2 until the automobile stops.

What total distance was traveled?

Answer in units of km.

- physics -
**drwls**, Saturday, September 15, 2012 at 1:26am
X1 = (1/2)a*19^2 = 234.7 m

Attained velocity:

Vmax = 1.3*19 = 24.7 m/s

X2 = 24.7 * 23 s = 568.1 m

X3 = (average speed)*(deceleration time)

= (Vmax/2)*(Vmax/a)= ___?

Add them up for total distance, and convert to km.

- physics -
**Eric**, Friday, January 31, 2014 at 5:03am
here is another way:

make it to X1,X2,X3

X1=(1/2)(1.3*19)(19)

X2=(1.3*19)(23)

X3=(1.3*19)/(0.9)(1/2)(1.3*19)

total distance = X1+X2+X3

## Answer This Question

## Related Questions

- Physics - An automobile accelerates from rest at 1.3 m/s 2 for 19 s. The speed ...
- Honors Physics - An automobile accelerates from rest at 1.3 m/s 2 for 19 s. The ...
- Physics - An automobile accelerates from rest at 2.4 m/s2 for 29 s. The speed is...
- physics - 1.The speed of a train increased from 15 mi/hr to 25 mi/hr in a ...
- Physics - An automobile and a truck start from rest at the same instant, with ...
- Physics - Could you please check my answers and units. Thanxs! 1. An automobile ...
- Science - I have some equations that I would like checked. Thanxs! 1. An ...
- Physics - Could you please check my answers and units.Also, help me with 3 and 4...
- 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...

More Related Questions