# Thread: Instantaneous Velocity

1. ## Instantaneous Velocity

If a ball is thrown straight up into the air with an initial velocity of $\displaystyle 50_ft/s$, its height in feet after t second is given by $\displaystyle y = 50 t - 16 t^2$. Find the average velocity for the time period beginning when $\displaystyle t = 1$ and lasting:

(i) $\displaystyle 0.1$ seconds =16.4 i.e., over the time interval $\displaystyle [1,1.1]$
(ii) $\displaystyle 0.01$ seconds=17.84

(iii) $\displaystyle 0.001$ seconds=17.984

Finally based on the above results, guess what the instantaneous velocity of the ball is when $\displaystyle t=1$$\displaystyle ? 2. Originally Posted by qbkr21 If a ball is thrown straight up into the air with an initial velocity of \displaystyle 50 ft/s, its height in feet after t second is given by \displaystyle y = 50 t - 16 t^2. Find the average velocity for the time period beginning when \displaystyle t = 1 and lasting: (i) \displaystyle 0.1 seconds =16.4 i.e., over the time interval \displaystyle [1,1.1] (ii) \displaystyle 0.01 seconds=17.84 (iii) \displaystyle 0.001 seconds=17.984 Finally based on the above results, guess what the instantaneous velocity of the ball is when \displaystyle t=1$$\displaystyle ?$
It looks like we are to use the formula:
$\displaystyle \bar{v} = \frac{\Delta y}{\Delta t}$

So for i) we have that
$\displaystyle y(1) = 50 \cdot 1 - 16 \cdot 1^2 = 34 \, ft$
$\displaystyle y(1.1) = 50 \cdot 1.1 - 16 \cdot 1.1^2 = 35.64 \, ft$

Thus
$\displaystyle \Delta y = y(1.1) - y(1) = 1.64 \, ft$

Thus
$\displaystyle \bar{v} = \frac{1.64 \, ft}{0.1 \, s} = 16.4 \, ft/s$

Similarly for ii) I get $\displaystyle \bar{v} = \frac{y(1.01) - y(1)}{0.01 \, s} = 17.84 \, ft/s$

And for iii) I get $\displaystyle \bar{v} = \frac{y(1.001) - y(1)}{0.001 \, s} = 17.984 \, ft/s$

It looks to me that the value of the instantaneous velocity at t = 1 s is 18 ft/s upward.

-Dan

3. Bingo!! Dan thanks alot you are a Master