# Thread: stuck on this problem

1. ## stuck on this problem

If a ball is thrown into the air with a velocity of 40ft./s its height in feet t seconds later is given by y=40t-16t^2.

a.)Find the average velocity for the time period beginning when t=2 and lasting

i.) 0.5 seconds ii.)0.1 seconds

iii.)0.05 seconds iv.)0.01 seconds

b.) Estimate the instantaneous velocity when t=2.

I am totally stuck I am not sure where to start I know. I know have to plug in the various times into the equation they gave. I also know that this problem has something to do wit-av. velocity=change in distance / change in time.

2. Originally Posted by DroMan
If a ball is thrown into the air with a velocity of 40ft./s its height in feet t seconds later is given by y=40t-16t^2.

a.)Find the average velocity for the time period beginning when t=2 and lasting

i.) 0.5 seconds ii.)0.1 seconds

iii.)0.05 seconds iv.)0.01 seconds
if the position function of an object is given by a function of time $\displaystyle f(t)$, then the average velocity between $\displaystyle t = a$ and $\displaystyle t=b$, for $\displaystyle a<b$ is given by:

$\displaystyle \text{Average velocity} = \frac {f(b) - f(a)}{b - a}$

this is really the formula for the slope between two points disguised, in case you need help remembering it.

for instance, for part (i)

let $\displaystyle f(t) = 40t - 16t^2$

then you want: $\displaystyle \frac {f(2.5) - f(2)}{2.5 - 2}$

b.) Estimate the instantaneous velocity when t=2.

I am totally stuck I am not sure where to start I know. I know have to plug in the various times into the equation they gave. I also know that this problem has something to do wit-av. velocity=change in distance / change in time.
you want to estimate the slope of a tangent to the curve at 2. in part (a) you kept taking the slope of points between 2 and something increasingly closer to 2. all these slopes get closer and closer to the value of the slope you are looking for. try to guess what these numbers are heading towards, and that is your estimate