# Math Help - Rate of Change

1. ## Rate of Change

How do you do this problem? I did (2.4-4.5)/(4-2) and got -.42. Then I did (4.5-7)/(20-15) and got -.5. I added -.42 and -.5 and got -.92 and divided that by 2 and my answer was: -.46 which makes no sense..?

2. Originally Posted by azncocoluver
How do you do this problem? I did (2.4-4.5)/(4-2) and got -.42.
Could you explain to us why you did that? I can see that "2.4" and "4.5" are the values for v(t) at times closes to t= 22 but where did "(4- 2)" come from? Well, actually, $\frac{2.4- 4.5}{4-2}= \frac{-2.1}{2}= -1.05$ not -.42 so I guess that was a typo. \frac{2.4- 4.5}{25- 20= \frac{-2.1}{5}= -.42. That looks to me like as good an answer as you can get from this data.

Then I did (4.5-7)/(20-15) and got -.5.
Well, I can see where you got both the numerator and denominator but those are for t= 15 and 20 and don't give any information about t= 22.

I added -.42 and -.5 and got -.92 and divided that by 2 and my answer was: -.46 which makes no sense..?
I must confess that why you are doing makes no sense. Why would you add those results and divide by 2?

The time given, t= 22, is not on the table but lies between t= 20 and t= 25. The average rate of change of v between those is $\frac{2.4- 4.5}{25- 20}= \frac{-2.1}{5}= -0.42$

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