Give an example using numbers please for this problem. Thanks!

A rocket is shot straight up into the air. If f(t) represents the height in feet of the rocket at time, t, seconds and g(t) is a constant function that represents the height of a platform, what situation can be described by (f+g) (t)? Include a domain of (f+g) (t).

1 answer

Sure! Let's create a specific example involving a rocket and a platform.

Example

Assume that the height of the rocket as a function of time \( t \) seconds is given by:

\[ f(t) = -16t^2 + 160t + 5 \]

In this function:

  • The term \(-16t^2\) represents the effect of gravity (in feet per second squared).
  • The term \(160t\) indicates that the rocket is initially launched with a velocity of 160 feet per second.
  • The constant \(5\) represents the initial height of the rocket above the ground at \( t = 0 \) seconds.

Next, let’s denote the height of the platform as a constant function represented by \( g(t) = 20 \). This means the platform is always at a height of 20 feet above the ground.

Now, we can express the combined height of both the rocket and the platform as:

\[ (f + g)(t) = f(t) + g(t) = (-16t^2 + 160t + 5) + 20 \]

Simplifying, we get:

\[ (f + g)(t) = -16t^2 + 160t + 25 \]

Interpretation

The function \( (f + g)(t) \) represents the total height from the ground of the rocket plus the height of the platform at any time \( t \). So if you want to know how high the rocket is above the platform at any time \( t \), you can subtract the height of the platform:

\[ \text{Height above platform} = f(t) = -16t^2 + 160t + 5 - 20 = -16t^2 + 160t - 15 \]

Example Domain

The domain of \( (f + g)(t) \) is determined by the context of the problem. For a rocket launched straight up, a reasonable domain could be from \( t = 0 \) to \( t = t_{\text{max}} \), where \( t_{\text{max}} \) is the time when the rocket reaches its peak height and begins to descend.

To find \( t_{\text{max}} \), we can calculate when the rocket hits the ground, or when \( f(t) = 0 \):

Solve \( -16t^2 + 160t + 5 = 0 \) using the quadratic formula:

\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-160 \pm \sqrt{(160)^2 - 4(-16)(5)}}{2(-16)} \]

Calculating this gives us two values for \( t \). However, for simplicity, let’s say the time to reach the peak height is around \( t = 10 \) seconds, so we can define our domain as:

\[ \text{Domain: } t \in [0, 10] \quad \text{(or until the rocket hits the ground if calculating those limits)} \]

Thus, the full situation can be summarized as follows:

  • The rocket is launched from a height of 5 feet with an initial velocity that lets it reach a peak height.
  • The platform stands at a height of 20 feet.
  • The combined function \( (f + g)(t) = -16t^2 + 160t + 25 \) gives the total height above the ground, and the domain is from \( t = 0 \) to when the rocket is back on the ground (or a specified duration).