initial velocity story problem

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needurhelp

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A ball is thrown straight up into the air with an initial velocity of 50 ft/s. Its height, in feet, after t seconds is given by y = 50t - 16t^2

Find the average velocity for the time period beginning when t = 2 and lasting:

a) 0.5sec
b) 0.1sec
c) 0.01sec

d) Finally based on the above results, guess what the instantaneous velocity of the ball is when t = 2

I' not good with story problem at all. Can someone help me start this problem?
 
You're given a formula for "y" in terms of "t", and various values for "t". Plug those values in for "t" in the formula, and find the corresponding values of "y".

You are asked for the average velocity over the intervals (from "t = 2" to "t = (whatever the other value is). To find the average velocity, find the distance covered between t = 2 and t = (whatever), and divide by the time interval (the "lasting" value).

For instance, part (a) asks for the average velocity on the interval t = 2 to t = 2.5. So plug "2" in for "t", and find the first value of "y". Then plug "2.5" in for "t", and find the second value of "y". Subtract the y-values to get the total distance, and then divide 0.5.

Parts (b) and (c) work the same way. Part (d) asks you to guess at what the value of the average would be as the "lasting" value gets smaller and smaller (that is, as the time "interval" gets closer and closer to being just "t = 2").

If you get stuck, please reply showing how far you have gotten. Thank you.

Eliz.

Edited to correct "y-values" to "velocity".
 
ok u said to plug in '2' for for t and solve for y. in part a i got:
y = 50(2)-16(2)^2
y = 36
y = 50(2.5)-16(2.5)^2
y = 25
average = (36+25)/2 = 30.5

is what im doing right?
 
Hello, needurhelp!

Let me baby-step through the concepts . . .

A ball is thrown straight up into the air with an initial velocity of 50 ft/s.
Its height, in feet, after \(\displaystyle t\) seconds is given by: \(\displaystyle \,y \:= \:50t\,-\,16t^2\)

Find the average velocity for the time period beginning when \(\displaystyle t\,=\,2\) and lasting:

. . a) 0.5 sec . . . b) 0.1 sec . . . c) 0.01 sec

d) Finally, based on the above results, guess what the instantaneous velocity
of the ball is when \(\displaystyle t\,=\,2\)
Click to expand...

Average velocity (speed) is always: \(\displaystyle \,\frac{\text{distance}}{\text{time}}\)

(a) At \(\displaystyle t\,=\,2\), the ball is: \(\displaystyle \,50(2)\,-\,16(2^2)\:=\:36\) feet above the ground.

At \(\displaystyle t\,=\,2.5\), the ball is: \(\displaystyle \,50(2.5)\,-\,16(2.5^2)\:=\:25\) feet above the ground.

The ball has travelled: \(\displaystyle \,25\,-\,36\:=\:-11\) feet (downward) in 0.5 seconds.

Its average velocity is: \(\displaystyle \,\frac{-11\text{ ft}}{0.5\text{ sec}} \:=\:-22\text{ ft/sec}\)


(b) At \(\displaystyle t\,=\,2\), the ball is 36 feet above the ground.

At \(\displaystyle t\,=\,2.1\), the ball is: \(\displaystyle 50(2.1)\,-\,16(2.1^2)\:=\:34.44\) feet above the ground.

It has moved: \(\displaystyle \,34.44\,-\,36\:=\:-1.56\) feet in 0.1 seconds.

Its average velocity is: \(\displaystyle \,\frac{-1.56\text{ ft}}{0.1\text{ sec}} \:=\:-15.6\text{ ft/sec}\)


(c) At \(\displaystyle t\,=\,2.01\), the ball is: \(\displaystyle 50(2.01)\,-\,16(2.01^2)\:=\:35.8584\) feet above the ground.

It has moved: \(\displaystyle \,35.8584\,-\,36\:=\:-0.1416\) in 0.01 seconds.

Its average velocity is: \(\displaystyle \,\frac{-0.1416\text{ ft}}{0.01\text{ sec}} \:=\:-14.16\text{ ft/sec}\)


(d) I think that it is a bit early to make a guess right now.
The velocity is certainly getting smaller, but is it heading toward 14 or 9 or 3?

Advice: Try a smaller time intervals, say, 0.001 second.
Find the average velocity for \(\displaystyle t\,=\,2.001\)
. . and we find that it is \(\displaystyle \,-14.016\text{ ft/sec}\)

Then I would guess that the instantaneous velocity at \(\displaystyle t\,=\,2\) is \(\displaystyle \,-14\text{ ft/sec.}\)
 
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