If a ball is thrown in the air with a velocity 46 ft/s, its height in feet t seconds later is given by y = 46t − 16t2.

(a) Find the average velocity for the time period beginning when t = 2 and lasting

0.5 second
0.1second
0.05 second
0.01 second

i tried plugging in 46(0.5)-16(0.5)^2 i got 19 and it was not the correct answer?? i don't know how else to to do it.

B)Estimate the instantaneous velocity when
t = 2
part b how would do you that?

1 answer

it says to start at t=2, not t=0.
So, you want the average velocity from 2 to 2.5

y(2) = 28
y(2.5) = 15
so, the average velocity is (15-28)/(2.5-2) = -13/.5 = -26

See

http://www.wolframalpha.com/input/?i=%28%2846t-16t^2%29-%2846*2-16*2^2%29%29%2F%28t-2%29+for+t%3D2.01%2C2.05%2C2.1%2C2.5

The instantaneous velocity at t=2 is

y'(t) = 46-32t
y'(2) = -18
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