Question

rockets two toy rockets are launched straight up into the air. the height, in feet, of each rocket at t seconds after launch is given by the polynomial equations shown.
rocket a: ( d_1 = -16t^2 + 122t )
rocket b: ( d_2 = -16t^2 + 84t )
a. write an equation to find the difference in height of rocket a and rocket b. interpret the parts of the equation in terms of the context.
the equation is
the difference in height by feet each second.
b. predict the difference in height after 5 seconds.
the difference in height after 5 seconds is feet.

Explanation:

Part a

Step 1: Find the difference equation

To find the difference in height between Rocket A (\(D_1 = -16t^2 + 122t\)) and Rocket B (\(D_2 = -16t^2 + 84t\)), we calculate \(D_1 - D_2\).
\[

$$\begin{align*} D_1 - D_2&= (-16t^2 + 122t) - (-16t^2 + 84t)\\ &= -16t^2 + 122t + 16t^2 - 84t\\ &= ( -16t^2 + 16t^2)+(122t - 84t)\\ &= 38t \end{align*}$$

\]
The equation for the difference in height is \(D_1 - D_2 = 38t\). The coefficient \(38\) means the difference in height increases by \(38\) feet each second (since \(t\) is in seconds, the rate of change of the difference with respect to time is \(38\) feet per second).

Step 2: Interpret the equation

The difference in height (of Rocket A and Rocket B) increases by \(38\) feet each second. So the first "Select Choice" for the difference equation is \(D_1 - D_2 = 38t\), the second "Select Choice" (for how the difference changes) is "increases", and the third "Select Choice" is \(38\).

Part b

Step 1: Substitute \(t = 5\) into the difference equation

We found the difference equation is \(38t\). Now substitute \(t = 5\) into it.
\[
38\times5 = 190
\]

Answer:

s:
a. The equation is \(D_1 - D_2 = 38t\). The difference in height increases by \(38\) feet each second.
b. The difference in height after 5 seconds is \(190\) feet.