1. word problem

the young folk decide to ride the (the Chaperones opt to watch from the comfort of the ground). At its maximum height, the ride reaches 65 m above the ground. When one boy reaches the top of the tower, his gum falls out of his mouth. The height of the gum can be given by the mathematical model: h=65-4.9t^2
where t is in seconds and h is measured in metres.

a. Find the average velocity of the gum on the intervals 2 < or equal t < or equal 3 and 2 < or equal t < 2.1
b. Find the instantaneous velocity when t=2

2. Originally Posted by william
the young folk decide to ride the (the Chaperones opt to watch from the comfort of the ground). At its maximum height, the ride reaches 65 m above the ground. When one boy reaches the top of the tower, his gum falls out of his mouth. The height of the gum can be given by the mathematical model: h=65-4.9t^2
where t is in seconds and h is measured in metres.

a. Find the average velocity of the gum on the intervals 2 < or equal t < or equal 3 and 2 < or equal t < 2.1
b. Find the instantaneous velocity when t=2
average velocity between two times $t_1$ and $t_2$ ...

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

instantaneous velocity at a specific time $t = a$

$v(a) = \lim_{t \to a} \frac{h(t) - h(a)}{t - a}$

3. Originally Posted by skeeter
average velocity between two times $t_1$ and $t_2$ ...

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

instantaneous velocity at a specific time $t = a$

$v(a) = \lim_{t \to a} \frac{h(t) - h(a)}{t - a}$
thanks, although i'm still not quite getting it can you give me a step further

4. On a sidenote:

For the first part of the problem, think a second for what "average" actually means.
Skeeter is saying (and correctly so) that he calculates the distance that the gum has fallen within the interval and then divide it with the difference in time.

I'd assume the solution to b would be the derivative of the function given - the derivative of a distance/time graph gives you velocity.

$
f(x) = 65 - 4.9t^2
$

$
f'(x) = 2 \cdot ( -4.9t ) = - 9.8t
$

$
f'(2) = -9.8 \cdot 2 = -19.6
$

5. Originally Posted by skeeter
average velocity between two times $t_1$ and $t_2$ ...

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

instantaneous velocity at a specific time $t = a$

$v(a) = \lim_{t \to a} \frac{h(t) - h(a)}{t - a}$
for a i am getting both negative for the first -24.5 and the second -2.01,correct?

i still can't get b

6. Originally Posted by william
for a i am getting both negative for the first -24.5 and the second -2.01,correct?
avg velocity over the interval [2,3] is -24.5 m/s

avg velocity over the interval [2,2.1] is -20.09 m/s

i still can't get b
$v(2) = \lim_{t \to 2} \frac{(65 - 4.9t^2) - 45.4}{t-2}$

$v(2) = \lim_{t \to 2} \frac{19.6 - 4.9t^2}{t-2}$

$v(2) = \lim_{t \to 2} \frac{4.9(4 - t^2)}{t-2}$

$v(2) = \lim_{t \to 2} \frac{4.9(2-t)(2+t)}{t-2}$

$v(2) = \lim_{t \to 2} \left[-4.9(2+t)\right] = \, \, ?$

7. Originally Posted by skeeter
avg velocity over the interval [2,3] is -24.5 m/s

avg velocity over the interval [2,2.1] is -20.09 m/s

$v(2) = \lim_{t \to 2} \frac{(65 - 4.9t^2) - 45.4}{t-2}$

$v(2) = \lim_{t \to 2} \frac{19.6 - 4.9t^2}{t-2}$

$v(2) = \lim_{t \to 2} \frac{4.9(4 - t^2)}{t-2}$

$v(2) = \lim_{t \to 2} \frac{4.9(2-t)(2+t)}{t-2}$

$v(2) = \lim_{t \to 2} \left[-4.9(2+t)\right] = \, \, ?$
i appreciate your help! i get -24.5 but for the second with

$
\frac{\delta h}{\delta t}=\frac{h(2.1)-h(2)}{2.1-2}=[65-4.9(2.1)^2]-[65-4.9(2)^2]=43.39-45.4=-2.01$

and for b/ i get 2

8. Originally Posted by william
i appreciate your help! i get -24.5 but for the second with

$
\frac{\delta h}{\delta t}=\frac{h(2.1)-h(2)}{\textcolor{red}{2.1-2}}=[65-4.9(2.1)^2]-[65-4.9(2)^2]=43.39-45.4=-2.01$

and for b/ i get 2
look at the denominator of the difference quotient ... you have to divide by 0.1

9. Originally Posted by skeeter
look at the denominator of the difference quotient ... you have to divide by 0.1
i understand, how do i finish that limit? i have not yet done limits like that, i am learning mathematics on my own, please pardon my slowness

10. Originally Posted by william
i understand, how do i finish that limit? i have not yet done limits like that, i am learning mathematics on my own, please pardon my slowness
then go and learn about limits ... find the limit when you are able.

11. Originally Posted by skeeter
then go and learn about limits ... find the limit when you are able.