Rocket Acceleration Calculation

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The discussion focuses on calculating the acceleration of a 1000 kg rocket during its initial 16 seconds of upward motion. The rocket motor provides a net constant acceleration, after which it coasts for 4 seconds before reaching an altitude of 5100 meters. Participants clarify that the rocket is not at rest at this altitude and propose using equations of motion to express velocity and height as functions of acceleration. By setting the height equation for the 4 seconds after motor shutdown equal to 5100 meters, an equation can be derived to solve for the acceleration. The mass of the rocket is deemed irrelevant to the calculation.
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A 1000 kg rocket is launched straight up. The rocket motor overcomes gravity to provide the rocket with a net constant acceleration upward for 16 seconds. The the motor shuts off. 4 seconds later the rocket's altitude is 5100 meters.
What is the rockets acceleration during the 1st 16 seconds?

0=v+-9.8 m/s^2 *4s = 39.2m/s but I'm off by a factor of 10 for the velocity at 5100 m
 
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What have you done so far?
 
jenduncan said:
0=v+-9.8 m/s^2 *4s = 39.2m/s but I'm off by a factor of 10 for the velocity at 5100 m

What exactly are you going for here? The assigment does not say that the rocket is at halt at 5100m.
 
Call the acceleration asked for "a". Then, for the first 16 seconds, v(t)= at and
h(t)= (a/2)t2 (v(t) is the velocity t seconds after lift off, h(t) is the height t seconds after lift off). From that you can find the velocity, v(16), and height h(16) when the rocket shuts off as a function of a.

Now the acceleration is just -g: v(t)= v16- gt and h(t)= h16+ v16- (g/2)t2, where v16 and h16 are the speed and height of the rocket when the rocket shut off. t is, of course, the time after the rocket shut off so set h(4)= 5100 and you will have an equation in a. Solve that equation.

Notice that the mass of the rocket is irrelelvant.
 
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