Find the average rate of change of the distance from home between 1 and 3 hours after leaving home.

Question

anonymous · Answer

attachment

inkyvoyd · Answer

$=(y_t-y_0)/(t-t_0)$

inkyvoyd · Answer

where $t_0=1$ and t=3

anonymous · Answer

what is this part =(yt−y0)

inkyvoyd · Answer

y_t is y at 3 hours, y_0 is y at 1 hour

inkyvoyd · Answer

165 and 55, respectively

I'm sorry I didn't offer you a clearer explanation, as I must go. You should get something around 110/2

anonymous · Answer

but i need a mph

inkyvoyd · Answer

the units are in miles per hour silly; 55 mph

anonymous · Answer

in the pack of my book it says 45mph im trying to figure out the work

anonymous · Answer

Question #2 What is another word for rate of change in this situation?

anonymous · Answer

Well, what's the rate of change of position?

Distance is measured in miles, and time in hours. The rate of change of position is given by
$$\frac{\Delta d}{\Delta t}=\frac{\text{miles}}{\text{hour}}$$
What does that unit correspond to?

anonymous · Answer

MPH

anonymous · Answer

Right, a unit of ...

anonymous · Answer

speed?

anonymous · Answer

Yes, or velocity (depending on the context).

anonymous · Answer

kk

anonymous · Answer

Siths in the back of my Algebra 2 textbook it says 45 mph but how do i show the work

anonymous · Answer

is there a formula

anonymous · Answer

Yeah, @inkyvoyd has already shown it. It's basically
$$\frac{\text{(distance at end of time interval)}-\text{(distance at start of time interval)}}{\text{(end time)}-\text{(start time)}}$$

anonymous · Answer

nvm im not thinking today when i looked at the answer in the back of the book it was for problem #41 and i was looking for number 40.  Thanks guys for both your help :D i really appreciate it.