Rate of Change for Ball Thrown: How to Use the Formula f(b)-f(a)/b-a

[Niaboc67]

Homework Statement

The Attempt at a Solution

As far as I know those are correct. Could I have some assistance on how to solve for the rate of change? I know the formula is f(b)-f(a)/b-a but not sure how to go about it

 

[SteamKing]

Homework Statement

The Attempt at a Solution

As far as I know those are correct. Could I have some assistance on how to solve for the rate of change? I know the formula is f(b)-f(a)/b-a but not sure how to go about it

What if f(t) = h(t)? Could you pick two times, t₁ = a and t₂ = b, and calculate the rate of change, knowing h(t) as a function of t?

 

[Brian T]

Homework Statement

The Attempt at a Solution

As far as I know those are correct. Could I have some assistance on how to solve for the rate of change? I know the formula is f(b)-f(a)/b-a but not sure how to go about it

That formula is saying take the change in height f(b) - f(a) and divide by the change in time b - a. Essentially, avg rate = total displavement over time for that displacement

 

[HallsofIvy]

Unfortunately most of the answers you give here are wrong. You are given that $h(t)= 56t- 16t^2= -16(t^2- (7/2)t)$. Completing the square gives $h(t)= -16(t- 7/2)^2+ 196$.

 

[mfb]

@HallsofIvy: You got a factor of 2 wrong for the last equation.

 

[Mark44]

I know the formula is f(b)-f(a)/b-a but not sure how to go about it

Use parentheses!
What you wrote is the same as $f(b) - \frac{f(a)}{b} - a$.

If you write a fraction such as the one above in single-line text, it has to be written as (f(b) - f(a))/(b - a).