Physics Help Forum Rocket Problem

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Jan 7th 2010, 01:20 PM #1 Junior Member   Join Date: Oct 2009 Posts: 6 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?
 Jan 7th 2010, 10:25 PM #2 Physics Team   Join Date: Feb 2009 Posts: 1,425 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
 Jan 8th 2010, 12:51 AM #3 Junior Member   Join Date: Oct 2009 Posts: 6 But where would the 325 m/s or the Supersonic speed Velocity play in?
 Jan 8th 2010, 04:49 AM #4 Physics Team   Join Date: Feb 2009 Posts: 1,425 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.

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