Question

question 4
height of a rocket
a toy rocket is shot from a platform 21 feet above the ground. the rocket flies into the air, reaches a
maximum height of 48 feet, then falls to the ground. the graph of h(t) below shows the height of this
rocket (in feet) t seconds after launch.
approximate the average rates of change over
each of the following time intervals. round to
two decimal places.
0 seconds to 30 seconds:
select an answer
30 seconds to 70 seconds:
select an answer
0 seconds to 70 seconds:
select an answer
complete the statement to explain the meaning of the rate of change in this situation.
on average, the height of the rocket is select an answer at an select an answer rate.
what do the values above tell you about the rate of change of h(t)?
the rate of change for h(t) is select an answer.

Explanation:

Step1: Recall the formula for average rate of change

The formula for the average rate of change of a function \(y = H(t)\) over the interval \([a,b]\) is \(\frac{H(b)-H(a)}{b - a}\).

Step2: Find \(H(0)\), \(H(30)\) and \(H(70)\)

From the graph, when \(t = 0\), \(H(0)=21\); when \(t = 30\), \(H(30) = 48\); when \(t=70\), \(H(70)=0\).

Step3: Calculate the average rate of change from \(t = 0\) to \(t = 30\)

Substitute \(a = 0\), \(b = 30\), \(H(a)=21\), \(H(b)=48\) into the formula \(\frac{H(b)-H(a)}{b - a}\).

$$ \frac{48 - 21}{30-0}=\frac{27}{30}=0.90 $$

Step4: Calculate the average rate of change from \(t = 30\) to \(t = 70\)

Substitute \(a = 30\), \(b = 70\), \(H(a)=48\), \(H(b)=0\) into the formula \(\frac{H(b)-H(a)}{b - a}\).

$$ \frac{0 - 48}{70 - 30}=\frac{-48}{40}=-1.20 $$

Step5: Calculate the average rate of change from \(t = 0\) to \(t = 70\)

Substitute \(a = 0\), \(b = 70\), \(H(a)=21\), \(H(b)=0\) into the formula \(\frac{H(b)-H(a)}{b - a}\).

$$ \frac{0 - 21}{70-0}=\frac{-21}{70}=- 0.30 $$

Answer:

0 seconds to 30 seconds: \(0.90\)
30 seconds to 70 seconds: \(-1.20\)
0 seconds to 70 seconds: \(-0.30\)