No, YOU show your work.
a) It begins to descend when V=0. Integrate the acceleration until time t when the integral is zero
b) Integrate the V(t) equation from (a) to get altitude Y vs t. Plug in the t you got in part (a)to get maximum height.
(c) Knowing the maximum height H, you can get the velocity after falling to earth from the equation
V = sqrt (2 a H)
(d) This should have been done while doing part (b)
This problem seems to be using an incorrect value for the acceleration "a" after 10 s. It should be 32 ft/s^2. Use the value they gave you, anyway. Also, for the first 10 seconds, the acceleration should be increasing, not decreasing, because the rocket's mass decreases with time during the time the engine is on. Whoever assigned the problem could use a refresher course in rockets
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.
Show all work. Thanks a lot!
3 answers
How high and how fast at 10 seconds first:
v = integral a dt + constant c
v(t) = 0 + 20 int dt e^-(t/2)
= 20 (-2) e^-(t/2) + c
= -40 e^-(t/2) + c
when t = 0, v = 0 so
0 = -40 + c
c = 40
so
v = 40(1-e^(-t/2) )for the first ten seconds
at ten seconds
v(10) = 40(1-.00674) = 39.7 m/s
h = int v dt + c2
h = 40 int (1 dt) -10 int (e^-(t/2))dt + c2
h = 40 t +80 e^(-t/2) + c2
when t = 0, h = 0
so c2 = -80
so
h = 40 t + 80 e^-(t/2) -80
at t = 10 seconds
h = 400 + .539 -80 = 321 m
SO
at ten seconds
Vo = 39.7 up
and
h = 321 meters up
Those are your initial conditions for the constant acceleration down phase. I think you can take it from there.
CHECK MY ARITHMETIC!!!
v = integral a dt + constant c
v(t) = 0 + 20 int dt e^-(t/2)
= 20 (-2) e^-(t/2) + c
= -40 e^-(t/2) + c
when t = 0, v = 0 so
0 = -40 + c
c = 40
so
v = 40(1-e^(-t/2) )for the first ten seconds
at ten seconds
v(10) = 40(1-.00674) = 39.7 m/s
h = int v dt + c2
h = 40 int (1 dt) -10 int (e^-(t/2))dt + c2
h = 40 t +80 e^(-t/2) + c2
when t = 0, h = 0
so c2 = -80
so
h = 40 t + 80 e^-(t/2) -80
at t = 10 seconds
h = 400 + .539 -80 = 321 m
SO
at ten seconds
Vo = 39.7 up
and
h = 321 meters up
Those are your initial conditions for the constant acceleration down phase. I think you can take it from there.
CHECK MY ARITHMETIC!!!
I mean feet, not meters.
Yes, acceleration is crazy. It should increase during the propulsion phase and it should be g, 32 ft/s^2 down, after the engine cuts out. I assume this is math class and certainly not a physics class :)
Yes, acceleration is crazy. It should increase during the propulsion phase and it should be g, 32 ft/s^2 down, after the engine cuts out. I assume this is math class and certainly not a physics class :)
1 answer