1. Develop an expression for its acceleration at a general time t . 2. If the ro
ID: 2012876 • Letter: 1
Question
1. Develop an expression for its acceleration at a general time t.
2. If the rocket starts with an initial velocity v0, derive the expression for its velocity at a general time t. [Be sure that the velocity reduces to vv0 at time t0].
3. If the rocket starts with an initial position y0, derive an expression for its position at a general time t. [Be sure that the position reduces to yy0 at time t0].
With help, I was able to correctly complete steps one and two (work shown below). However, I'm stuck at step three. I realize the position equation is just the integral of the velocity, but I can't seem to find the correct solution.
Ugh... I typed out all my work and it cut it off. I don't feel like typing it all again, so I'll just state that
Explanation / Answer
F=ma, m is constant, so we can just have a be equal to the force at that time/m.
F0e^(-Bt)/m-g for a (so I agree)
integrate it for b and I agree with you again.
well, now we just integrate it again (same process as how you got 2?)
y0+v0t-gt^2/2+middle term. I'll integrate it with work for you
integral F0/(Bm)(1-e^(-Bt) dt
the first part is a constant so it's really F0/(Bm)*integral (1-e^(-Bt) dt
F0/(Bm)*e^(-Bt)/B+t+constant
but we want it to be 0 at 0, so it's F0/(Bm)*((e^(-Bt)-1)/B+t)
y0+v0t-gt^2/2+F0/(Bm)*((e^(-Bt)-1)/B+t)
and at 0 everything goes to 0 besides y0, so it works.
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