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Kinematics: finding acceleration

  • Thread starter kapatter
  • Start date
  • #1
2
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I've been trying to figure this question out, and am now near on to pulling my hair out.
A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 seconds, then the motor stops. The rocket altitude 20 seconds after launch is 5100 meters. You can ignore any effects of air resistance. What was the rocket's acceleration during the first 16 seconds?

In my panick for a solution, one thing I tried was this:
v = u+at,
= 0 +16a
= 16a
distance travelled during first 16 second
s= ut +(1/2) at^2
= 0 + (1/2) at^2 = (1/2) a 20^2 = 200a

total distance travelled after 20 second = 5100 m

5100 = 200a + 16a x4 -(1/2) x9.8 x4^2
264a = 5178.4
a = 19.61 m/sec^2 ....but obviously, this is incorrect. If anybody has any suggestions at all, I would be really grateful. Thanks!
 

Answers and Replies

  • #2
berkeman
Mentor
56,139
6,170
You're on the right track, but making small errors in your math. First, in your initial calculation of s, you say that you want to calculate s after 16 seconds, and then plug in t=20 into the equation.

So fix that, and then make it clear what the velocity and acceleration are doing during the two time periods to give you the total.
 
  • #3
957
0
I think you have a good start. The distance traveled during the last 4 seconds if I'm thinking right should be treated kinematically as the distance traveled with the Vi being the velocity after 16 secs subject to the decelleration due to gravity. That could be a delta Y. The other Y achieved during the first 16 seconds is 5100 less delta Y.

So keep on keeping on but develop an all at once closed form solution, that can be evaluated. This way logical vs computational errors can be eassily tested.
 

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