A little cal help please average velocity?

Question

The statement is
If a ball is thrown in the air with a velocity 42 ft/s, its height in feet t seconds later is given by y = 42t - 16t2

What to find
Find the average velocity for the time period beginning when t = 2 and lasting 0.5 second

I am not sure how to set this up can someone please show me?

Answer 1

average = [f(2.5) -f(2)]/2.5-2)
= (5-20)/.5 = -30ft/sec

Answer 2

The function giving the height of the ball after t seconds:

s(x) = 42t - 16t^2

The first derivative of this function is the function giving the instantaneous velocity of the ball at any t:

s'(x) = v(x) = 42 - 32t

When t = 2, the velocity equals v(2) = 42 - 32(2) = 42 - 64 = -22 ft/s

0.5 seconds later, at t=2.5, the velocity equals v(2.5) = 42 - 32(2.5) = 42 - 80 = -38 ft/s

The average velocity during this interval is therefore (-22 + -38)/2 = -60/2 = -30 ft/s

Answer 3

You can actually do this without calculus. Average velocity is the distance traveled divided by the time it took. So, all you need is the two positions of the ball.

42(2) - 16(4) = height after 2 seconds = 20 feet

42(2.5) - 16(6.25) = height 0.5 seconds later = 5 feet

Average velocity = (20-5)/(0.5) = 30 feet/second

Answer 4

y=42t-16t^2
height after 2 seconds
y(2)=84-64=20ft
height after 2.5 seconds
y(2.5)=105-100=5ft
distance travelled downward in 0.5 seconds=20-5=15ft
Average velocity during the interval=15/0.5=30ft/s.downward

Answer 5

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Answer 6

V=U+aT

a=-32.15fps(squared)

U=42fps

T=2,2.5

V1= 42+(-32.15)2 (at time 2 seconds)

V1= -22.3 fps (coming back down again)

V2=42+(-32.15)2.5 (at time 2.5 seconds)

V2=-38.375 fps

Va= (22.3+38.375)/2 (Va=average velocity)

Va= 30.3375 fps (a rough average velocity)

PS, i hope i didn't just do your homework.