1. ## average velocity

I have a problem that I am stuck on, The problem states: If a ball is thrown straight up into the air with an initial velocity of 90 ft/s, its height in feet after t seconds is given by y = 90t - 16t^2. Find the average velocity for the time period begining when t = 1 and lasting:
a) 0.1 sec. over the time interval [1, 1.1]
b.) 0.01 sec.
c.) 0.001 sec.
Finally based on the above results, guess what the instantaneous velocity of the ball when t = 1.

I set it up( 90(1.1) - 16(1)^2)/0.1
I did the example in the book and this one is a little different because the ball is going up, the other question has the ball dropped from a certain height. I did take t=1 and plug it in the equation and got 99-16 = 83/.01 and this was not correct. I tried using a form of the difference quotient but not very successful. Let me know if one of these ideas is the right way to go. It is possible I just made a simple mistake and I have made the same mistake over and over again and did not catch on until I asked for help, so if you could please give me a hint, I would appreciate it!
Thank you,
Keith Stevens

2. Originally Posted by kcsteven
I have a problem that I am stuck on, The problem states: If a ball is thrown straight up into the air with an initial velocity of 90 ft/s, its height in feet after t seconds is given by y = 90t - 16t^2. Find the average velocity for the time period begining when t = 1 and lasting:
a) 0.1 sec. over the time interval [1, 1.1]
b.) 0.01 sec.
c.) 0.001 sec.
Finally based on the above results, guess what the instantaneous velocity of the ball when t = 1.

I set it up( 90(1.1) - 16(1)^2)/0.1
Why this?

For part (a)

y(1)=90-16 ft
y(1.1)=90(1.1)-16(1.1)^2=(90-16)+(9-16*0.21) ft

so the average velocity = [y(1.1)-y(1)]/0.1=(9-16*0.21)/0.1=56.4 ft/s

RonL

3. ## average velocity

I still do not get it, sorry. In the second part of the equation:
y(1.1) = 90(1.1) -16(1.1)^2 = (90-16) + (9-16*0.21) ft.
I can not figure out where the nine comes from and also the 0.21.
If you have time would you explain how you arrived at those numbers.=.
Thank you!!!!!!!
Keith Stevens

4. Originally Posted by kcsteven
I still do not get it, sorry. In the second part of the equation:
y(1.1) = 90(1.1) -16(1.1)^2 = (90-16) + (9-16*0.21) ft.
I can not figure out where the nine comes from and also the 0.21.
If you have time would you explain how you arrived at those numbers.=.
Thank you!!!!!!!
Keith Stevens
90(1.1)=90(1+0.1)=90+9

16(1.1)^2=16(1.21)=16(1+0.21)=16+16*0.21

RonL

5. Originally Posted by kcsteven
I have a problem that I am stuck on, The problem states: If a ball is thrown straight up into the air with an initial velocity of 90 ft/s, its height in feet after t seconds is given by y = 90t - 16t^2. Find the average velocity for the time period begining when t = 1 and lasting:
a) 0.1 sec. over the time interval [1, 1.1]
b.) 0.01 sec.
c.) 0.001 sec.
Finally based on the above results, guess what the instantaneous velocity of the ball when t = 1.

I set it up( 90(1.1) - 16(1)^2)/0.1
I did the example in the book and this one is a little different because the ball is going up, the other question has the ball dropped from a certain height. I did take t=1 and plug it in the equation and got 99-16 = 83/.01 and this was not correct. I tried using a form of the difference quotient but not very successful. Let me know if one of these ideas is the right way to go. It is possible I just made a simple mistake and I have made the same mistake over and over again and did not catch on until I asked for help, so if you could please give me a hint, I would appreciate it!
Thank you,
Keith Stevens
Average velocity = (distance travelled) / (time spent)

a) Between t = 1 sec and t = 1.1 sec
y(t) = 90t -16(t^2)
y(1) = 90*1 -16(1)^2 = 74 ft
y(1.1) = 90*1.1 -16(1.1^2) = 79.64 ft.
So, distance travelled = 79.64 -74 = 5.64 ft
Time spent = 0.1 sec
Hence, average velocity = 5.64 /0.1 = 56.4 ft/sec ------***

b) Between t = 1 sec and t = 1.01 sec
y(t) = 90t -16(t^2)
y(1) = 90*1 -16(1)^2 = 74 ft
y(1.01) = 90*1.01 -16(1.01^2) = 74.5784 ft.
So, distance travelled = 74.5784 -74 = 0.5784 ft
Time spent = 0.01 sec
Hence, average velocity = 0.5784 /0.01 = 57.84 ft/sec ------***

c) Between t = 1 sec and t = 1.001 sec
y(t) = 90t -16(t^2)
y(1) = 90*1 -16(1)^2 = 74 ft
y(1.001) = 90*1.001 -16(1.001^2) = 74.057984 ft.
So, distance travelled = 74.057984 -74 = 0.057984 ft
Time spent = 0.001 sec
Hence, average velocity = 0.057984 /0.001 = 57.98 ft/sec ------***

"Finally based on the above results, guess what the instantaneous velocity of the ball when t = 1."

We can see that as the time spent after t = 1 second gets smaller and smaller, the average velocity is getting higher slower and slower, tending to become 58 ft/sec at t=1 sec.
Therefore, the instantanous velocity at t=1 second is 58 ft/sec . ------answer.

--------------------------------------
You are maybe studying Calculus now.
In Calculus, the instantaneous rate is the value of the first derivative of the function.
y(t) = 90t -16t^2
y'(t) = 90 -32t
At t=1.
y'(1) = 90 -32(1) = 58 ft/sec ----------same as above.

6. ## average velocity

Thank you, that was an excellent explaination. I sat and looked at that problem for an hour yesterday and earlier this morning I looked at it again and realized it was the time interval that was giving me the problem, I could not grasp the time interval for each problem [1,1.1], [1,1.01], [1,1.001] yesterday and after a good nite sleep it just seemed too easy. I guess sometimes you have to put it away and concentrate on another subject or take a break.
Thank you for your excellent explaination!
Keith Stevens

7. Originally Posted by kcsteven
Thank you, that was an excellent explaination. I sat and looked at that problem for an hour yesterday and earlier this morning I looked at it again and realized it was the time interval that was giving me the problem, I could not grasp the time interval for each problem [1,1.1], [1,1.01], [1,1.001] yesterday and after a good nite sleep it just seemed too easy. I guess sometimes you have to put it away and concentrate on another subject or take a break.
Thank you for your excellent explaination!
Keith Stevens
Yes, that is problem-solving!
Put it away for a while, then come back to attack it again with a fresh mind. On the same approach or from another angle.

8. Hello, Keith!

If a ball is thrown straight up into the air with an initial velocity of 90 ft/s,
its height in feet after t seconds is given by: $y\:=\:90t - 16t^2$

Find the average velocity for the time period begining when t = 1 and lasting:

a) 0.1 sec . . b) 0.01 sec . . c) 0.001 sec

Finally based on the above results,
guess the instantaneous velocity of the ball when t = 1.

Work out each average velocity.

We find that: . $y(1) \:= \:90(1) - 16(1^2) \:= \:74$

On $[1,\,1.1]\!:\;y(1.1) \:=\:90(1.1) - 16(1.1^2) \:=\:79.64$
Then: . $v\:=\:\frac{79.64-74}{0.1} \:=\:\frac{5.664}{0.1} \:=\:56.4$

On $[1, 1.01]\!:\;y(1.001)\:=\:90(1.01) - 16(1.01^2) \:=\:74.5784$
Then: . $v\:=\:\frac{74.5784- 74}{0.01} \:=\:\frac{0.5784}{0.01} \:=\:57.84$

On $[1,\,1.001]\!:\;y(1.001)\:=\:90(1.001) - 16(1.001^2) \:=\:74.057984$
Then: . $v \:=\:\frac{74.057984 - 74}{0.001} \:=\:\frac{0.057984}{0.001} \:=\:57.984$

I would guess that $v \rightarrow 58.$

9. ## average velocity

Thank you for your explaination your's always seems to be the clearest. I appreciate you taking the time to reply!
Thank you,
Keith Stevens