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!
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!!!
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 :)
To find the time at which the rocket begins to descend, we need to find when the acceleration changes from positive to negative.
a) The rocket begins to descend when the acceleration changes from positive to negative. In this case, the acceleration changes at t = 10 seconds, as given by the piecewise function. So, the rocket begins to descend at t = 10 seconds.
To find the height the rocket reaches, we need to integrate the acceleration function twice with respect to time.
b) To find the height, we first integrate the acceleration function once to get the velocity function:
v(t) = ∫(20e^(-t/2)) dt = -40e^(-t/2) + C1
Here, C1 is the constant of integration.
Next, we integrate the velocity function to get the position function:
s(t) = ∫(-40e^(-t/2) + C1) dt = 80e^(-t/2) + C1t + C2
Here, C2 is the constant of integration.
Now, we can find the value of C1 and C2 using the initial condition given that the initial velocity of the rocket is zero. So, v(0) = 0.
0 = -40e^(0) + C1
C1 = 40
Therefore, the velocity function becomes:
v(t) = -40e^(-t/2) + 40
Now, using the position function, we can find the maximum height by finding the value of s(t) when v(t) = 0. This is because the rocket reaches the maximum height when its velocity becomes zero.
-40e^(-t/2) + 40 = 0
e^(-t/2) = 1
- t/2 = 0
t = 0
Substituting t = 0 into the position function:
s(t) = 80e^(-t/2) + 40t + C2
s(0) = 0 + 0 + C2
C2 = 0
Therefore, the position function becomes:
s(t) = 80e^(-t/2) + 40t
Hence, the rocket reaches a maximum height of 80 feet.
c) To find the velocity when the rocket impacts the Earth, we need to find the velocity as t approaches infinity. As t approaches infinity, e^(-t/2) approaches 0. Therefore, the velocity when the rocket impacts the Earth is given by:
v(t) = -40e^(-t/2) + 40
v(∞) = -40(0) + 40
v(∞) = 40 feet per second
d) The position of the rocket with respect to time for t > 10 seconds is given by the formula obtained earlier:
s(t) = 80e^(-t/2) + 40t, for t > 10 seconds.
Please note that the provided formulas apply under the assumption that the rocket starts from rest at t = 0.
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