My Math Forum Acceleration as rocket fuel burns

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December 6th, 2014, 01:59 PM   #1
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Is this corrrect? Also, I'm not sure how to do e and f
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Last edited by Omnipotent; December 6th, 2014 at 02:12 PM.

December 7th, 2014, 06:26 AM   #2
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Quote:
 Originally Posted by Omnipotent Is this corrrect? Also, I'm not sure how to do e and f
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
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Quote:
 Originally Posted by Omnipotent But isn't the "jerk" the rate of change of acceleration aka the derivative of a(t)?
Yes, that is what you have tried to find and other than attempting to differentiate a constant ($I$) you have not done the part of the derivative you can do - this is part (c)

CB

Last edited by skipjack; December 7th, 2014 at 11:37 PM.

December 7th, 2014, 10:35 AM   #5
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Quote:
 Originally Posted by CaptainBlack Yes, that is what you have tried to find and other than attempting to differentiate a constant ($I$) you have not done the part of the derivative you can do - this is part (c) CB
I see, thank you!
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
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Quote:
 Originally Posted by CaptainBlack Yes, that is what you have tried to find and other than attempting to differentiate a constant ($I$) you have not done the part of the derivative you can do - this is part (c) CB
corrected question c),

I*k / (Mr + Mf - kt)

December 7th, 2014, 07:20 PM   #7
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Quote:
 Originally Posted by Omnipotent corrected question c), I*k / (Mr + Mf - kt)
Yes

CB

December 7th, 2014, 07:50 PM   #8
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Quote:
 Originally Posted by Omnipotent 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?
The integral in (d) is of the form:$$\int -\frac{I}{M_r+M_f-kt}\; dt=\frac{I}{k} \ln(M_r+M_f-kt)+C$$

So putting in the limits of integration gives:$$\int_0^t -\frac{I}{M_r+M_f-kt}\; dt=\frac{I}{k} \ln(M_r+M_f-kt)-\frac{I}{k} \ln(M_r+M_f)=\frac{I}{k} \ln\left(\frac{M_r+M_f-kt}{M_r+M_f}\right)$$

CB

Last edited by CaptainBlack; December 7th, 2014 at 07:55 PM.

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