Solve Kinematics: Rocket Accel 1350.0kg, 10.2s, 2210m

[emilinus]

I just need help getting started with this question.

A 1350.0 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 10.2 s, then the motor stops. The rocket altitude 26.9 s after launch is 2210.0 m. You can ignore any effects of air resistance. What was the rocket's acceleration during the first 10.2 s?

total distance is 2210.0 m
acceleration during 10.2 s and 26.9 s is -9.81 m/s^2
the acceleration during t=0 s and t=10.2 s is cst.

 

[olgranpappy]

you have a kinematic equation which relates position velocity and acceleration, something like:
<br /> x = \frac{1}{2}at^2+v_0 t+x_0\;.<br />
a is accleration, v_0 is initial velocity, x_0 is initial position, x is position at time t. First of all, what are x_0, and v_0?

 

[emilinus]

but say x_0 and v_0 are both 0...you're still left with two variables...

ie x = 52.2a well that's for the first section t=0 to t=10.2

 

[olgranpappy]

but say x_0 and v_0 are both 0...you're still left with two variables...

you are given both 'x' and 't' in the third sentence of the question. plug them in and solve for 'a'.

 

[emilinus]

that would give the acceleration during t=10.2 s and t=26.9 s...which is different from the acceleration during t=0 s to t=10.2 s

 

[olgranpappy]

oh... right, I see.

Well, you can work in two steps. For the first 10.2 seconds you already know that x(t) = 1/2 at^2

so you can find the position at t=10.2 (in terms of a) and the velocity at t=10.2 (in terms of a). Call these things x_1 and v_1, respectively.

Next, you can use the "free fall" kinematic equations (i.e., the usual equation with a=-g=-9.8m/s^2) to do the rest of the problem; just use v_1 and x_1 where you would normally have used v_0 and x_0. And use (t-10.2) where you would normally have used t.

P.S. Draw a picture of what's going on, that should help a bit and help you organize what is happening to the rocket.