Calculus Rocket Problem (Derivatives)

In summary, the conversation discusses a model rocket being launched vertically upward and finding its position and instantaneous velocity and acceleration at any time t. The first derivative of the position function is used to find the velocity, while the second derivative is used to find the acceleration. It is determined that the acceleration is a constant -32 ft/s^2, indicating a constant force of gravity acting on the rocket. The conversation also mentions finding the maximum height of the rocket, which can be done by setting the velocity equation equal to 0 and solving for time. The result is used to find the maximum height using the original position equation.
  • #1
frumdogg
18
0

Homework Statement



A model rocket is launched vertically upward. If it's position (height) y above the ground in feet as a function of time t in seconds is given by:

Homework Equations



y = f(t) = 192t - 16t^2

The Attempt at a Solution



1. Find the instantaneous velocity and acceleration at any time t.

Now this is my first calculus course on my long academic journey, but velocity is the first derivative and acceleration is the second.

Velocity -

y' = 192-32t

That was easy, but it is the second derivative (acceleration) that is stumping me:

y'' = -32

How is that possible.. negative acceleration the entire time?

Thank you for your assistance.
 
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  • #2
That's not much of a rocket! If y= 192t- 16t2, then the "rocket" provides an (instantaneious!) initial 192 ft/sec speed and then shuts down. The only force acting on it is that due to gravity. Are you sure that the y= 192t- 16t2 isn't from thepoint at which the rocket shuts off?
 
  • #3
That is the equation, verbatim from the homework problem.

One of the other questions is (and I have done the work for it):

2. How high will the rocket reach?

192 - 32t = 0

-32t = -192
-32 = -32

total time = 6 seconds

192(6) - 16(36) = 576 feet
 
  • #4
That's the correct approach. Plug your critical point back into your equation, and you find the Y_max!
 
  • #5
Thanks roco, but is the second derivative (acceleration) correct at -32? It just does not seem logical to me.
 

1. What is the "Calculus Rocket Problem"?

The Calculus Rocket Problem is a mathematical problem that involves using derivatives to find the maximum height a rocket can reach based on its initial velocity and acceleration.

2. How do you set up the problem?

To set up the problem, you first need to identify the given variables such as the initial velocity and acceleration of the rocket. Then, you can use the formula for position as a function of time (x(t)) and take its derivative to find the velocity (x'(t)). Finally, you take the derivative of velocity (x''(t)) to find the acceleration and set it equal to 0 to solve for the time when the rocket reaches its maximum height.

3. What are the steps for solving the problem?

The steps for solving the Calculus Rocket Problem are as follows:
1. Identify the given variables
2. Use the formula x(t) to find the velocity function x'(t)
3. Take the derivative of x'(t) to find the acceleration function x''(t)
4. Set x''(t) equal to 0 and solve for t to find the time when the rocket reaches its maximum height.
5. Substitute the value of t into x(t) to find the maximum height of the rocket.

4. What are the key concepts involved in solving this problem?

The key concepts involved in solving the Calculus Rocket Problem are derivatives and optimization. Derivatives are used to find the rate of change of a function, which is crucial in determining the velocity and acceleration of the rocket. Optimization is used to find the maximum height by setting the acceleration function equal to 0 and solving for the time when the rocket reaches its peak.

5. What are some real-life applications of this problem?

The Calculus Rocket Problem has various real-life applications, such as designing and optimizing rocket trajectories for space exploration, calculating the maximum height of a projectile for sports like basketball and football, and determining the maximum height of a rollercoaster. It is also used in physics, engineering, and other fields that involve motion and optimization.

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