Question

Help with only part D!

Answer 1

Time
(in minutes)
x 0 10 20 30 40
Car 1
(in km)
y 40 30 20 10 0
Car 2
(in km)
y 40 32.68292 26.70432 21.81936 17.828


Part A: What does the x-intercept of the function for car 1 represent? (2 points)

Part B: What does the y-intercept of the function for car 2 represent? (2 points)

Part C: What is the domain of the functions for car 1 and car 2? (2 points)

Part D: What is the average rate of change from x = 10 to x = 30 for the function representing the motion for car 2? What does the value of this average rate of change represent? (4 points)

Answer 2

i thought it was 20.

Answer 3

@shamil98 @SolomonZelman @AllTehMaffs @KattGirl14 @nincompoop

Answer 4

What is the average rate of change of car 2 from time = 10-30 min?

Answer 5

yes.

Answer 6

Do you have the function that represents car 2's motion?

Answer 7

Cool, thanks.

So, the 20 that you got is the change in displacement of car 1 from x=10 to x=30, but you really want the average *rate of change of car 2 - rates deal with time. The rate of change in displacement is
\[\frac{\Delta \text{position}}{\Delta \text{time}}=\frac{\Delta y}{\Delta x}\]

So, first, what's the rate of change in displacement of car 2 from time=10min to time=20 min?

Then, what's average the rate of change in displacement of car 2 from time=20min to time=30min?

Answer 8

\[\frac{\Delta y}{\Delta x}=\frac{y_f - y_i}{x_f-x_i}\]

Answer 9

i got 10 is that right?

Answer 10

@AllTehMaffs

Answer 11

How dd you get that?

in between x=10min and x=20min
\[\frac{\Delta y}{\Delta x} = \frac{ 26.70432km - 32.68292km}{20 min - 10 min} = \frac{-5.9786km}{10 min}\]
and in between x=20 min and x=30 min
\[\frac{21.81936km - 26.70432km}{30 min - 20 min}\]

Then the average rate of change is the average of those two - it should be negative, and less than 1

Answer 12

oh ok...

Answer 13

How did you get 10?

Answer 14

10 - 20 = 10
20 - 30 = 10...lol heheh...

Answer 15

-4.88496/10

Answer 16

That's the change in time, not the rate of change of displacement. A rate implies the change in some variable with respect to time- in this case, the rate of change is the change in position with respect to time.

Answer 17

That's the rate fo change of position in between time=20 min and time=30 min.

Now you want the average rate of change in between time=10min and time=30min, so you average the rate of change between time=10 min , time=20min and time = 20min, time=30min

Answer 18

I made this slightly more complicated than it needed to be, in order to show that the rates of change in between those two fragments of time was different. You can get the same answer by just using initial time=10 min and final time = 30 min and the corresponding positions.
\[\frac{21.81936km - 32.68292km}{30 min - 10 min}\]
gives the same answer as the average of the other two rates.


In any case, what does this average rate represent?

Answer 19

i got -1634145.009032

Answer 20

that seems tad high, don't you think?

Answer 21

oh nvm...i got .543178.

Answer 22

much more reasonable! (and it's correct)

Can you say what are the units are on that number?

Answer 23

-10.86356/20

Answer 24

but it's -10.86356 what? seconds? furlongs? miles?

And 20 what? Joules? parsecs? cabbages?

Answer 25

20 minutes and -10.86356 km?

Answer 26

most definitely! So that means you had
\[\frac{-10.86356 km}{20min} = -.543178 \ km/min\]
the units are kilometers per minute.

Do those units tell you anything about what that rate might represent? It's position per time - what other units that you use daily are in position per time?

Answer 27

A tank yu...wery helpful :D

Answer 28

^_^ Welcome.

Answer 29

Can you say what that rate is, however? (just to humor me ^_^)

Answer 30

which rate 10? or -.543178

Answer 31

The rate is the thing with units km/min (-.543178 km/min) - "10min" isn't a rate, it's a difference in time.

so yeah, what does -.543178 km/min represent?

Answer 32

da rate of chicken i mean change.

Answer 33

It's actually the average speed of the car over that time interval.

Answer 34

oh...interesting...so whats the average rate of change?

Answer 35

That's what it is - the average rate of change in position versus time is the average speed of the car is equal to -.543178 km/min

Answer 36

oh ok...u got me scared just now lol...

Answer 37

^_^