Question

If a ball is thrown straight up into the air with an initial velocity of 40 ft/s, it height in feet after t second is given by y=40t−16t2. Find the average velocity for the time period begining when t=2 and lasting
(i) 0.5 seconds

(ii) 0.1 seconds

(iii) 0.01 seconds

Finally based on the above results, guess what the instantaneous velocity of the ball is when t=2.

Answer 1

This is a derivative problem

Answer 2

the derivative of position is velocity

Answer 3

Edutopia, that is not true. It's supposed to guide one to a better understanding of the difference between the average and the instantaneous velocity.

Nick, think back to the definition of the average velocity (total distance travelled divided by time spent traveling that distance) and apply it to those time intervals.

Answer 4

well the maximum height occurs at

\[t = \frac{-40}{2 \times -16}\]

so the max hight occurs when t = 1.25 seconds...

so when t = 2

the height is zero when

\[40t - 16t^2 = 0 ... or...8t(5 - 2t) = 0\]

so the height is zero when t = 0 or 2.5 seconds...

so from this we can see the postion of the ball

ok... so find y when t =2

y = 40(2) - 16(2)^2

y = 26... so the ball is moving down and is 16 ft above the group.

now find y when t = 2.5

y = 40(2.5) - 16(2.5)^2

y = 0

so the ball has moved 16 ft in 0.5 seconds...

average speed = distance traveled/time taken

so

Average speed = 16/0.5 = 32 ft/sec

same process for for t = 2.1 find y

then the change in distance over time taken will given average speed.

Hope this makes sense...

Answer 5

what you are being asked to do is to find the slope of the secant... to get average speed, but as the points are getting closer together the average speed is getting closer to the instantaneous velocity.