Calc
posted by Michael on .
A rocket is launched with an initial velocity of zero, and with acceleration in feet per second per second defined by:
( 20e^(t/2), for 0 <= t <= 10 seconds)
a(t) = (
( 16, for t > 10 seconds)
a) At what time does the rocket begin to descend?
b) How high does the rocket reach?
c) What is the velocity when the rocket impacts the earth?
d) Write a formula for the position of the rocket with respect to time for t > 10 seconds.
Thanks!

I believe both drwls and I did this problem not long ago.
It is kind of wacky.
The acceleration should increase, not decrease, while the engine is on for the first ten seconds because the mass decreases as fuel is used up.
The acceleration down during the ballistic phase with the engine off should be 32 ft/s^2, gravity, not 16.
Never the less I will get you started.
a = d^2h/dt^2 = 20 e^t/2
v = dh/dt = c  40 e^(t/2)
since v = 0 at t = 0, c = 40 and
v = 40 (1e^(t/2))
Now we can get v at t = 10
v = 39.7 ft/s at t = 10
how high is it at t = 10?
h = c2 + 40 t + 80 e^(t/2)
h = 0 when t = 0 so c2 = 80
h = 40 t +80(e^(t/2) 1)
now when t = 10
h = 400 + 80 (.0061) = 320
so NOW we have a brand new problem at t = 10 seconds
a = 16
Vo = 39.7 ft/s
Ho = 320 ft
Now
v = vo  at
at the top, v = 0
0 = 39.7  16 t
t = 2.48 s
SO part a)
t+the original ten seconds = 12.48 s
h = Ho + Vo t + (1/2) a t^2
h = 320 + 39.7 * 2.48 8 *2,48^2
h = 320 + 98.5  49.2
So part b)
h = 369 feet at the top.
c it falls from 369 feet
v = 0  32 t
369 = (1/2)(32) t^2
369 = 16 t^2
t = 4.80 seconds falling
SO
Part c
v at ground = 32(4.8) =  154 ft/s
I can not do part d because they had g wrong, 16 instead of 32 during the ballistic phase.
If it were 32, as it should be, then
h = Ho + Vo (t10) + (1/2)(32) (t10)^2
Ho is 320
Vo is 39.7 If I did the first ten seconds right
I used t10 because I assume they mean total t since liftoff, not time after ten seconds.