# Thread: Help needed with a simple velocity problem

1. ## Help needed with a simple velocity problem

Hi all. I need some kind person to explain to me what exactly to do in this simple problem.

A ball is thrown into the air with a velocity of 40ft/sec. It's height in feet after t seconds is y=40t-16(t^2). Find the average velocity for the time period beginning when t=2 and lasting .05 seconds.

Also find the instantaneous velocity when t=2.

2. Originally Posted by kep84
Hi all. I need some kind person to explain to me what exactly to do in this simple problem.

A ball is thrown into the air with a velocity of 40ft/sec. It's height in feet after t seconds is y=40t-16(t^2). Find the average velocity for the time period beginning when t=2 and lasting .05 seconds.

Also find the instantaneous velocity when t=2.

The average velocity between $\displaystyle t_1$ and $\displaystyle t_2>t_1$ is:

$\displaystyle \bar{v}=\frac{y(t_2)-y(t_1)}{t_2-t_1}$

RonL

3. Originally Posted by kep84
Hi all. I need some kind person to explain to me what exactly to do in this simple problem.

A ball is thrown into the air with a velocity of 40ft/sec. It's height in feet after t seconds is y=40t-16(t^2). Find the average velocity for the time period beginning when t=2 and lasting .05 seconds.

Also find the instantaneous velocity when t=2.

The instantaneous velocity at $\displaystyle t$ is:

$\displaystyle v(t)=\left. \frac{dy}{dt}\right|_t =40-32t$

RonL

4. Originally Posted by CaptainBlack
The average velocity between $\displaystyle t_1$ and $\displaystyle t_2>t_1$ is:

$\displaystyle \bar{v}=\frac{y(t_2)-y(t_1)}{t_2-t_1}$

RonL

ummm could you go a bit further please? i don't know what i'm supposed to do with this.

Originally Posted by CaptainBlack
The instantaneous velocity at $\displaystyle t$ is:

$\displaystyle v(t)=\left. \frac{dy}{dt}\right|_t =40-32t$

RonL
i've never seen this before...
$\displaystyle \left. \frac{dy}{dt}\right|_t$

5. Originally Posted by kep84
i've never seen this before...
$\displaystyle \left. \frac{dy}{dt}\right|_t$
It means you evaluate the derivative at $\displaystyle t$.

6. Originally Posted by kep84
ummm could you go a bit further please? i don't know what i'm supposed to do with this.

i've never seen this before...
$\displaystyle \left. \frac{dy}{dt}\right|_t$
to use other notations you may be familiar with.

Let $\displaystyle y = f(t)$

Average velocity from a point a to a point b is: $\displaystyle \frac {f(b) - f(a)}{b - a}$

the "points" here are values of t

The instantaneous velocity at time t is: $\displaystyle f'(t)$

so you need to find $\displaystyle f'(2)$ for the second part

7. thanks for the help so far guys. can someone please just do the problem and show me how they are plugging in the numbers so i can just see exactly how it's done one time?

8. Originally Posted by kep84
Hi all. I need some kind person to explain to me what exactly to do in this simple problem.

A ball is thrown into the air with a velocity of 40ft/sec. It's height in feet after t seconds is y=40t-16(t^2). Find the average velocity for the time period beginning when t=2 and lasting .05 seconds.
we want the average velocity between $\displaystyle t = 2$ and $\displaystyle t = 2.05$

Let $\displaystyle y = f(t)$

$\displaystyle \Rightarrow \mbox {Average Velocity} = \frac {f(2.05) - f(2)}{2.05 - 2}$

$\displaystyle = \frac {14.76 - 16}{0.05}$

$\displaystyle = -24.8$

Also find the instantaneous velocity when t=2.
$\displaystyle f(t) = 40t - 16t^2$

$\displaystyle \Rightarrow f'(t) = 40 - 32t$

Now, the instantaneous velocity when $\displaystyle t = 2$ is given by:

$\displaystyle f'(2) = 40 - 32(2) = -24$

9. Originally Posted by Jhevon
we want the average velocity between $\displaystyle t = 2$ and $\displaystyle t = 2.05$

Let $\displaystyle y = f(t)$

$\displaystyle \Rightarrow \mbox {Average Velocity} = \frac {f(2.05) - f(2)}{2.05 - 2}$

$\displaystyle = \frac {14.76 - 16}{0.05}$

$\displaystyle = -24.8$

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

$\displaystyle \Rightarrow f'(t) = 40 - 32t$

Now, the instantaneous velocity when $\displaystyle t = 2$ is given by:

$\displaystyle f'(2) = 40 - 32(2) = -24$
thx a bunch jhevon and the rest of you.