# 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 $t_1$ and $t_2>t_1$ is:

$
\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 $t$ is:

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

RonL

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

$
\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 $t$ is:

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

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

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

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...
$
\left. \frac{dy}{dt}\right|_t
$
to use other notations you may be familiar with.

Let $y = f(t)$

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

the "points" here are values of t

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

so you need to find $f'(2)$ 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 $t = 2$ and $t = 2.05$

Let $y = f(t)$

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

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

$= -24.8$

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

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

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

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

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

Let $y = f(t)$

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

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

$= -24.8$

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

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

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

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