Posted by **John** on Saturday, August 28, 2010 at 6:11pm.

Practice problem:

A rocket sled for testing equipment under large accelerations starts at rest and accelerates according to the expression:

a= (2.8 m/s^3)t + (3.9 m/s^2)

How far does the rocket move in the time interval t=0 to t=0.81 s?

This is just a practice problem to help me get an equation for my actual problem.

I know acceleration is the 2nd derivative and so I feel like I should integrate this equation (twice?) but my integration abilities are sorely lacking; especially as it regards to including units.

If I integrate how does that affect m/s^3 and m/s^2? Is it a definite integral from 0 to 0.81 s? While I feel I'm on something of the right track here I'm unsure.

Thanks in advance for the help!

- physics (mechanics) -
**drwls**, Saturday, August 28, 2010 at 8:10pm
Yes, two integrations are required.

V(t) = integral of a(t) dt from t = 0 to t

V(t) = 1.4 t^2/2 + 3.9t MINUS the value of the same function at t=0 (which is zero), so

V(t) = 1.4 t^2/2 + 3.9t

X(t) - X(0) = integral of V(t) from t = 0 to t

X(0) = 0

X(t) = 1.4 t^3/6 + 1.95 t^2

If t is in seconds, X will be in meters

## Answer this Question

## Related Questions

- Physics - An ice sled powered by a rocket engine starts from rest on a large ...
- Physics - An ice sled powered by a rocket engine starts from rest on a large ...
- Physics - An ice sled powered by a rocket engine starts from rest on a large ...
- physics - An ice sled powered by a rocket engine starts from rest on a large ...
- Physics - An ice sled powered by a rocket engine starts from rest on a large ...
- physics - An ice sled powered by a rocket engine starts from rest on a large ...
- physics - An ice sled powered by a rocket engine starts from rest on a large ...
- physics - A rocket-driven sled running on a straight, level track has been used...
- Classical Mechanics Physics - Consider a rocket in space that ejects burned fuel...
- physics - A rocket, initially at rest on the ground, accelerates straight upward...

More Related Questions