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 − 16t2.
(a) Find the average velocity for the time period beginning when t = 2 and lasting
With the following answers:
(i) 0.5 second.
(ii) 0.1 second.
(iii) 0.05 second.
(iv) 0.01 second.
(b) Estimate the instantaneous velocity when t = 2.
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
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.
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- 64-16h^2-64h +64 / h
=-16h^2 - 24h/ n
= h (-16h - 24)/ h
= -16h - 24
Next she plugs in the values.
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.
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?