Chapter 14.4, Problem 49AYU

Instantaneous Velocity on the Moon Neil Armstrong throws a ball down into a crater on the moon. The height $s$ (in feet) of the ball from the bottom of the crater after $t$ seconds is given in the following table:

(a) Find the average velocity from $t=1$ to $t=4$ seconds.

(b) Find the average velocity from $t=1$ to $t=3$ seconds.

(c) Find the average velocity from $t=1$ to $t=2$ seconds.

(d) Using a graphing utility, find the quadratic function of best fit.

(e) Using the function found in part (d), determine the instantaneous velocity at $t=1$ second.

To determine

To find: Instantaneous Velocity on the Moon Neil Armstrong throws a ball down into a crater on the moon. The height $s$ (in feet) of the ball from the bottom of the crater after $t$ seconds is given in the following table:

a. Find the average velocity from $t=1$ to $t=4$ seconds.

Answer to Problem 49AYU

Solution:

a. $-23\frac{1}{3}\text{ft/sec}$

Explanation of Solution

Given:

Calculation:

a. The average velocity from $t=1$ to $t=4$ is,

$\frac{\text{Δs}}{\text{Δt}}=\frac{s\left(4\right)-s\left(1\right)}{4-1}$

$s\left(4\right)=917;s\left(1\right)=987$

$\frac{\text{Δs}}{\text{Δt}}=\frac{s\left(4\right)-s\left(1\right)}{4-1}=\frac{917-987}{3}=\frac{-70}{3}=-23\frac{1}{3}\text{ft/sec}$

To determine

To find: Instantaneous Velocity on the Moon Neil Armstrong throws a ball down into a crater on the moon. The height $s$ (in feet) of the ball from the bottom of the crater after $t$ seconds is given in the following table:

b. Find the average velocity from $t=1$ to $t=3$ seconds.

Answer to Problem 49AYU

Solution:

b. $-21\text{ft/sec}$

Explanation of Solution

Given:

Calculation:

b. The average velocity from $t=1$ to $t=3$ is,

$\frac{\text{Δs}}{\text{Δt}}=\frac{s\left(3\right)-s\left(1\right)}{3-1}$

$s\left(3\right)=945;s\left(1\right)=987$

$\frac{\text{Δs}}{\text{Δt}}=\frac{s\left(3\right)-s\left(1\right)}{3-1}=\frac{945-987}{2}=\frac{-42}{2}=-21\text{ft/sec}$

To determine

To find: Instantaneous Velocity on the Moon Neil Armstrong throws a ball down into a crater on the moon. The height $s$ (in feet) of the ball from the bottom of the crater after $t$ seconds is given in the following table:

c. Find the average velocity from $t=1$ to $t=2$ seconds.

Answer to Problem 49AYU

Solution:

c. $-18\text{ft/sec}$

Explanation of Solution

Given:

Calculation:

c. The average velocity from $t=1$ to $t=2$ is,

$\frac{\text{Δs}}{\text{Δt}}=\frac{s\left(2\right)-s\left(1\right)}{2-1}$

$s\left(4\right)=969;s\left(1\right)=987$

$\frac{\text{Δs}}{\text{Δt}}=\frac{s\left(2\right)-s\left(1\right)}{2-1}=\frac{969-987}{1}=-18\text{ft/sec}$

To determine

To find: Instantaneous Velocity on the Moon Neil Armstrong throws a ball down into a crater on the moon. The height $s$ (in feet) of the ball from the bottom of the crater after $t$ seconds is given in the following table:

d. Using a graphing utility, find the quadratic function of best fit.

Answer to Problem 49AYU

Solution:

d. $-2.631t^2–10.269t+999.933$

Explanation of Solution

Given:

Calculation:

d. Using a graphing utility, find the quadratic function of best fit.

The quadratic function is $y=-2.631t^2–10.269t+999.933$.

To determine

To find: Instantaneous Velocity on the Moon Neil Armstrong throws a ball down into a crater on the moon. The height $s$ (in feet) of the ball from the bottom of the crater after $t$ seconds is given in the following table:

e. Using the function found in part $\left(d\right)$, determine the instantaneous velocity at $t=1$ second.

Answer to Problem 49AYU

Solution:

e. $–15.531\text{ft/sec}$

Explanation of Solution

Given:

Calculation:

e. The instantaneous velocity of $t_1$ is the derivative $s’\left(t_1\right)$: that is,

$s\left(t_1\right)=\underset{t\to t_1}{\lim }\frac{s\left(t\right)-s\left(t_1\right)}{t-t_1}$

$=\underset{t\to t_1}{\lim }\frac{\left(-2.631t2–10.269t+999.933\right)-\left(-2.631t_1^2-10.269t_1+999.933\right)}{t-t_1}$

$=\underset{t\to t_1}{\lim }\frac{-2.631\left(t^2-t_1^2\right)-10.269\left(t-t_1\right)}{t-t_1}$

$=\underset{t\to t_1}{\lim }\frac{-2.631\left[\left(t+t_1\right)\left(t-t_1\right)\right]-10.269\left(t-t_1\right)]}{t-t_1}$

$=\underset{t\to t_1}{\lim }\frac{\left(t-t_1\right)(-2.631\left(t+t_1)-10.269\right)}{t-t_1}$

$=\underset{t\to t_1}{\lim }(-2.631\left(t+t_1)-10.269\right)$

$=-2.631\left(t_1+t_1\right)–10.269\text{ft/sec}$

Replace $t_1$ by $t$. the instantaneous velocity of the ball at time $t$ is,

$s\left(t\right)=-2.631\left(2t\right)–10.269\text{ft/sec}$

The instantaneous velocity at $t=1$ is,

$s\left(1\right)=-2.631\left(2\left(1\right)\right)–10.269=-15.531\text{ft/sec}$