December 6th, 2014, 01:59 PM  #1 
Member Joined: Sep 2014 From: Cambridge Posts: 64 Thanks: 4 
Is this corrrect? Also, I'm not sure how to do e and f
Last edited by Omnipotent; December 6th, 2014 at 02:12 PM. 
December 7th, 2014, 06:26 AM  #2 
Senior Member Joined: Jan 2012 From: Erewhon Posts: 245 Thanks: 112  a) Looks OK b) $kt$ is the mass of fuel (in kg) burnt in the first $t$ seconds of the engine burn. . c) $I$ is a constant so $I'=0$, and you need to actually take the derivative that is left. d) You have: $\frac{dv}{dt}=a(t)$, so: $$ v(t)=v_0+\int_0^t a(t)\; dt $$ I will leave you to do the integral. e) Just plug $t=75$ in when you have the answer to d). f) You have $\frac{ds}{dt}=v(t)$, where $s$ is the distance down range. CB Last edited by CaptainBlack; December 7th, 2014 at 06:28 AM. 
December 7th, 2014, 08:59 AM  #3 
Member Joined: Sep 2014 From: Cambridge Posts: 64 Thanks: 4 
But isn't the "jerk" the rate of change of acceleration aka the derivative of a(t)?
Last edited by skipjack; December 7th, 2014 at 11:36 PM. 
December 7th, 2014, 09:43 AM  #4  
Senior Member Joined: Jan 2012 From: Erewhon Posts: 245 Thanks: 112  Quote:
CB Last edited by skipjack; December 7th, 2014 at 11:37 PM.  
December 7th, 2014, 10:35 AM  #5  
Member Joined: Sep 2014 From: Cambridge Posts: 64 Thanks: 4  Quote:
Sorry to bug you one more time but, for question c) I got: a'(t) = (I)(Mr+Mf kt)' / (Mr+Mf kt)^2 Is this correct? Also, for question d), I'm not sure how to find the integral of such an equation, could you help me out on how to approach it?  
December 7th, 2014, 01:26 PM  #6 
Member Joined: Sep 2014 From: Cambridge Posts: 64 Thanks: 4  
December 7th, 2014, 07:20 PM  #7 
Senior Member Joined: Jan 2012 From: Erewhon Posts: 245 Thanks: 112  
December 7th, 2014, 07:50 PM  #8  
Senior Member Joined: Jan 2012 From: Erewhon Posts: 245 Thanks: 112  Quote:
\int \frac{I}{M_r+M_fkt}\; dt=\frac{I}{k} \ln(M_r+M_fkt)+C $$ So putting in the limits of integration gives:$$ \int_0^t \frac{I}{M_r+M_fkt}\; dt=\frac{I}{k} \ln(M_r+M_fkt)\frac{I}{k} \ln(M_r+M_f)=\frac{I}{k} \ln\left(\frac{M_r+M_fkt}{M_r+M_f}\right) $$ CB Last edited by CaptainBlack; December 7th, 2014 at 07:55 PM.  

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