SOLUTION: KMST NOTE: I removed the unneeded space. The young folk decide to ride the Terror Tower (the Chaperones opt to watch from the comfort of the ground!). At its maximum height, t

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Question 1118286: KMST NOTE:
I removed the unneeded space.

The young folk decide to ride the Terror Tower (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.
[Marks: a: 4; b: 6]
a. Find the average velocity of the gum on the intervals 2 <_ t <_3 and 2 <_ t <_2.1
b. Find the instantaneous velocity when t=2.
all questions please
Found 3 solutions by Alan3354, josgarithmetic, KMST:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Why all the space?

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
If your model has to be h=65-9t%5E2, then the gum hits the ground for h=0;

65-9t%5E2=0
65=9t%5E2
t%5E2=65%2F9
t=sqrt%2865%29%2F3
t=2.687, or to the nearest whole second, 3seconds.

You can find the asked average speeds in the other given time intervals.
%28h%5B2%5D-h%5B1%5D%29%2F%28t%5B2%5D-t%5B1%5D%29

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The height of the gum can be given by h=65-4.9t%5E2 , where t is in seconds and h is measured in metres.

a. To find the average velocity of the gum on the intervals 2+%3C=+t+%3C=3 and 2+%3C=+t+%3C=2.1
you calculate height of the gum at t=2, t=2.1, and t=3,
and use those values to calculate average velocity.

For t=2 , h=65-4.9%2A2%5E2=65-4.9%2A4=65-19.6=45.4 .
For t=2.1 , h=65-4.9%2A2.1%5E2=65-4.9%2A4.41=65-21.609=43.391 .
For t=3 , h=65-4.9%2A3%5E2=65-4.9%2A9=65-44.1=20.9 .

In the interval 2+%3C=+t+%3C=3 , the height of the gum changes from 45.4 m to 20.9 m in 1 second.
The average downwards velocity in the interval 2+%3C=+t+%3C=2.1 is
%2845.4m-20.9m%29%2F%283s-2s%29%22=%2224.5m%2F%221+s%22%22=%22%2224.5+m+%2F+s%22 .

In the interval 2+%3C=+t+%3C=2.1 , the height of the gum changes from 45.4 m to 43.391 m in 0.1 second.
The average downwards velocity in the interval 2+%3C=+t+%3C=2.1 is
%2845.4m-43.391m%29%2F%282.1s-2s%29%22=%222.009m%2F%220.1+s%22%22=%22%2220.09+m+%2F+s%22

b. Knowing about derivatives, you would know that the instantaneous velocity when t=2 is the value of the derivative of h with respect to t when t=2.
The derivative of the function h=65-4.9t%5E2 with respect to t is
dh%2Fdt=-4.9%2A2t=-9.8t , and its value for t=2 is -9.8%2A2=-19.6 .
The derivative is negative because height decreases with time,
but you could say the instantaneous downwards velocity when t=2 is %2219.6+m+%2F+s%22 .
If you have not been taught about derivatives as functions, maybe you were expected to calculate it as a limit:
lim%28t-%3E2%2C%28h%28t%29-h%282%29%29%2F%28t-2%29%29%22=%22lim%28t-%3E2%2C%2865-4.9t%5E2-%2865-4.9%2A4%29%29%2F%28t-2%29%29%22=%22lim%28t-%3E2%2C-4.9%28t%5E2-4%29%2F%28t-2%29%29%22=%22lim%28t-%3E2%2C-4.9%28t%2B2%29%28t-2%29%2F%28t-2%29%29%22=%22lim%28t-%3E2%2C-4.9%28t%2B2%29%29%22=%22-4.9%282%2B2%29%29%22=%22-4.9%2A4%22=%22-19.6 .