
New Member
Average Velocity
Again, I'm sure this doesn't go here but it's a problem I got in my Calculus class, so I'm posting it here.
I have the following question:
If a ball is thrown in the air with a velocity 34 ft/s, its height in feet t seconds later is given by y = 34t − 16t^{2}.
(a) Find the average velocity for the time period beginning when t = 2 and lasting
With the following answers:
(i) 0.5 second.
38 ft/s
(ii) 0.1 second.
31.6 ft/s
(iii) 0.05 second.
30.8 ft/s
(iv) 0.01 second.
30.16 ft/s
(b) Estimate the instantaneous velocity when t = 2.
30 ft
But they're all wrong. Here's how I worked it out:
Avg Velocity= (change in)y/(change in) x= y (2+ h) y (2)/(2+h)  2
Then, I worked all that through and got = 16h 30
I plugged the value of h in each time.
AV= 16 (.5)30= 38ft/sec
AV= 16 (.1)  30= 31.6 ft/sec
etc, etc.
Can negative numbers work with this? Do I need to just change the signs? Or did I screw up bad somewhere in the math? Also, is there an easier way to figure this out?
I can only submit this one more time, so I need to make SURE I have the right answer this time.

New Member
The example I watched had the following:
If a ball is thrown into the air with a velocity of 40 ft/s, its height in feet t seconds later is given by y= 40t  16t^2.
Find the average velocity for the time period beginning when t=2 and lasting
i) 0.5 seconds
First she used a formula for average velocity.
t=2 to t= 2 + h
(change in)y/(change in)t = y (2 + h) y(2)/ (2+h)  (2)
= 40 (2+ h)  16 (2 + h)^2  [40 (2) 16 (2)^2] / h
=80 + 40h  16 (4 + h^2 + 4h)  80 + 64/ h
= 40h 6416h^264h +64 / h
=16h^2  24h/ n
= h (16h  24)/ h
= 16h  24
Next she plugs in the values.
h= .5
A.V.= 16 (.5)  24= 32ft/s
&so on and so on.
But this was the example I was given from the text for our class.

New Member
Thanks! Your way seems much more simple.
So, should the answer for 0.1 be 32.4?
See this threw me off because I saw the velocity was going from 34, to 26, to 32.4?
Then, for 0.05= 33.2?
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