Question

1. when a flea jumps off the back of a horse onto a dog, its height off the ground (in centimeters) after t seconds is given by: $h(t)=370t - 490t^{2}+170$. the flea reaches a maximum height after 0.4 seconds and lands on the dog after 0.9 seconds. (a) what is the height of the dog where the flea lands? include units in your answer. (b) compute the net change in the fleas height between the peak at 0.4 seconds and the landing at 0.9 seconds. include units in your answer. (c) compute the average rate of change in the fleas height between the peak at 0.4 seconds and the landing at 0.9 seconds. include units in your answer.

Explanation:

Step1: Find height when flea lands

We know the flea lands at $t = 0.9$ seconds. Substitute $t=0.9$ into $H(t)=370t - 490t^{2}+170$.
$H(0.9)=370\times0.9 - 490\times(0.9)^{2}+170$
$=333-490\times0.81 + 170$
$=333 - 396.9+170$
$=106.1$ centimeters.

Step2: Find height at peak

Substitute $t = 0.4$ into $H(t)=370t - 490t^{2}+170$.
$H(0.4)=370\times0.4-490\times(0.4)^{2}+170$
$=148 - 490\times0.16+170$
$=148-78.4 + 170$
$=239.6$ centimeters.

Step3: Compute net - change in height

Net change is $H(0.9)-H(0.4)$.
$106.1 - 239.6=-133.5$ centimeters.

Step4: Compute average rate of change

The formula for average rate of change of a function $y = f(x)$ from $x=a$ to $x = b$ is $\frac{f(b)-f(a)}{b - a}$. Here, $a = 0.4$, $b = 0.9$, $f(t)=H(t)$.
Average rate of change $=\frac{H(0.9)-H(0.4)}{0.9 - 0.4}=\frac{- 133.5}{0.5}=-267$ centimeters per second.

Answer:

(a) $106.1$ centimeters
(b) $-133.5$ centimeters
(c) $-267$ centimeters per second