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 (1-e^-(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 (.006-1) = 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 (t-10) + (1/2)(-32) (t-10)^2
Ho is 320
Vo is 39.7 If I did the first ten seconds right
I used t-10 because I assume they mean total t since liftoff, not time after ten seconds.
How do you get "-1" in h(t) equation?
a) To find out when the rocket begins to descend, we need to determine when its acceleration changes from positive to negative. From the given information, the rocket's acceleration is 20e^(-t/2) for 0 <= t <= 10 seconds and -16 for t > 10 seconds.
At t = 10 seconds, the acceleration changes from 20e^(-t/2) to -16. Therefore, at t = 10 seconds, the rocket begins to descend.
b) To find out how high the rocket reaches, we need to integrate the rocket's velocity function with respect to time. Since the initial velocity is zero, the velocity function can be found by integrating the rocket's acceleration function.
For 0 <= t <= 10 seconds:
v(t) = ∫(20e^(-t/2)) dt
v(t) = -40e^(-t/2) + C1
For t > 10 seconds:
v(t) = ∫(-16) dt
v(t) = -16t + C2
To find C1 and C2, we need initial conditions. Since the rocket starts with an initial velocity of zero, we can set v(0) = 0.
At t = 0 seconds:
v(0) = -40e^(-0/2) + C1
0 = -40(1) + C1
C1 = 40
So the velocity function becomes:
v(t) = -40e^(-t/2) + 40 for 0 <= t <= 10 seconds
v(t) = -16t + C2 for t > 10 seconds
Now, to find the maximum height, we need to integrate the velocity function with respect to time again.
For 0 <= t <= 10 seconds:
h(t) = ∫(-40e^(-t/2) + 40) dt
h(t) = -80e^(-t/2) + 40t + C3
For t > 10 seconds:
h(t) = ∫(-16t + C2) dt
h(t) = -8t^2 + C2t + C4
To find C3 and C4, we again need initial conditions. As the rocket starts from rest, h(0) = 0.
At t = 0 seconds:
h(0) = -80e^(0/2) + 40(0) + C3
0 = -80(1) + C3
C3 = 80
So the height function becomes:
h(t) = -80e^(-t/2) + 40t + 80 for 0 <= t <= 10 seconds
h(t) = -8t^2 + C2t + C4 for t > 10 seconds
To find the maximum height, we can use the height function for 0 <= t <= 10 seconds since the rocket descends after that.
Therefore, the rocket reaches a maximum height of -80e^(-10/2) + 40(10) + 80 feet.
c) To find the velocity when the rocket impacts the earth, we can substitute t = 10 seconds into the velocity function for t > 10 seconds:
v(10) = -16(10) + C2
v(10) = -160 + C2
Since the rocket impacts the earth, its velocity is zero. Therefore, the velocity when the rocket impacts the earth is:
0 = -160 + C2
C2 = 160
So the velocity function when the rocket impacts the earth is:
v(t) = -16t + 160 for t > 10 seconds.
d) To write a formula for the position of the rocket with respect to time for t > 10 seconds, we need to integrate the velocity function:
h(t) = ∫(-16t + 160) dt
h(t) = -8t^2 + 160t + C4
Since the rocket impacts the earth, the position when the rocket impacts the earth is zero. Therefore, we have:
0 = -8(10)^2 + 160(10) + C4
0 = -800 + 1600 + C4
C4 = -800
So the position function for t > 10 seconds is:
h(t) = -8t^2 + 160t - 800.
Note: The units for the position, velocity, and acceleration in this explanation are in feet and seconds, as stated in the problem.