Physics Help Forum (http://physicshelpforum.com/physics-help-forum.php)
-   Kinematics and Dynamics (http://physicshelpforum.com/kinematics-dynamics/)
-   -   Rocket Problem (http://physicshelpforum.com/kinematics-dynamics/3912-rocket-problem.html)

 link88 Jan 7th 2010 12:20 PM

Rocket Problem

The rocket starts from rest at t = 0 and travels straight up. Its height above the ground as a function of time can be approximated by s = bt^2 + ct^3, where b and c are constants. At t = 10s, the rockets velocity and acceleration are v = 229 m/s and a = 28.2 m/s^2. Determine the time at which he rocket reaches supersonic speed (325 m/s). What is its altitude when that occurs.

Im not sure what to do with the s =bt^2 + ct^3 equation, and also don't know why I would need b and c as constants?

 physicsquest Jan 7th 2010 09:25 PM

a and b are constants given as part of the problem. You have to find their values to solve it.

Differentiate (take the derivative of) the s equation with respect to t

This will give you a v equation as v = ds/st = ...........
where v is the velocity.

In this equation, plug in the given values of v and t.

This will give you one equation relating a and b.

Now differentiate the v equation with respect to t

This will give give you the accln say A = dv/dt.

Plug in the value of t and A.

This gives you the second equation relating a and b.

Solve these two equations simultaneously to get a and b.

Put these back into the v equation and then find the time.

Use this value of time in the s equation to get altitude

 link88 Jan 7th 2010 11:51 PM

But where would the 325 m/s or the Supersonic speed Velocity play in?

 physicsquest Jan 8th 2010 03:49 AM

That is what was mentioned in this line

Put these back into the v equation and then find the time.

In the v eqn, put v = 325 m/s and solve for t since you know a and b.

Then use the t you found above in the s eqn to get s.

 All times are GMT -7. The time now is 02:45 AM.