Asked by B <3

*VELOCITY OF A CAR* Suppose the distance s (in feet) covered by a car moving along a straight road after t sec is given by the function s = f(t) = 2t^2 + 48t.

a. Calculate the average velocity of the car over the time intervals [20,21], [20, 20, 20.1], and [20, 20.01].

b. Calculate the (instantaneous) velocity of the car when t = 20.

c. Compare the results of part (a) with that of part (b).

*Can someone please help me? I have been trying to do my homework for some time now and I am stuck on this problem. I tried to work it out, but I did not understand what to substitute in for “t”. I used the average velocity formula and did f(t+h)-f(t) divided by h …. And I got: 4t + 48. Somehow, I think this is way wrong! Can you tell me where I went wrong here and how to do this problem!?? Pleaseee!

any help would be great!!! :)
Answered by bobpursley
The instantaneous velocity is indeed 4t+48

However, the average velocity is the average of the instaneous velocties at two times. Example, in a, times 20,and times 20.1
v(20)=4t+48=128
v(20.1)=128.4

average: 128.2
Answered by Anonymous
do you have the dike too?
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