# Calculus Average Velocity

## Homework Statement

A ball is thrown into the air a velocity of 49 ft/s. Its height in feet after t seconds is given by y=49t-10t^2.

A. Find the average velocity for the time period beginning when t=3 and lasting
0.01 s:
0.005 s:

B. Estimate the instantaneous velocity when t=3.

Integrals

## The Attempt at a Solution

I'm not exactly sure if I'm approaching this problem at the right angle...

A) 0.01s

1/ [(3.01) -3] ∫ from 3 to 3.01 49t-10t^2 dx

[1/0.01] ∫ from 3 to 3.01 49t-10t^2 dx

[1/0.01] ∫ from 3 to 3.01 (49t^2)/2 - (10t^3)/3 <------ integrated

[1/0.01] [(49(3.01)^2)/2 - (10(3.01)^3)/3] - [1/0.01] [ 49(3)^2)/2 - (10(3)^3)/3

substitution

Same steps for 0.005s

B) I'm not sure how to approach this problem.

eumyang
Homework Helper
I'm not exactly sure if I'm approaching this problem at the right angle...
No, you don't want to take an integral. You're given a position function, and integrating it doesn't give you the velocity function. What you want to do is to use the average rate of change formula (from x = a to x = b):
$$\frac{f(b) - f(a)}{b - a}$$
So, given the position function s(t) = -10t2 + 49t, evaluate
$$\frac{s(3.01) - s(3)}{3.01 - 3}$$

B) I'm not sure how to approach this problem.
Do you know the formula for finding a derivative at a point (I'm talking about the one with the limit)?

Do you know the formula for finding a derivative at a point (I'm talking about the one with the limit)?

Actually, I don't know the formula for finding a derivative at a point.

eumyang
Homework Helper
Actually, I don't know the formula for finding a derivative at a point.
Sure you do:
$$f'(a) = \lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h}$$

Sure you do:
$$f'(a) = \lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h}$$

Thanks. :D