If a ball is thrown into the air with a velocity of 48 ft/s, its height in feet t seconds later is given by y = 48t − 16t2.
(a) Find the average velocity for the time period beginning when t = 2 and lasting for each of the following.
(i) 0.5 seconds
ft/s
(ii) 0.1 seconds
ft/s
(iii) 0.05 seconds
ft/s
(iv) 0.01 seconds
ft/s
(b) Estimate the instantaneous velocity when t = 2.
ft/s
y(t) = 48t − 16t^2 (note notation)
i)
y(2.5) =48(2.5)-16(6.25) = 20
y(2) = 48(2)-16(4) = 32
20-32 = -12 ft in .5 s
-12/.5 = -24
ii)
y(2.1) = 48*2.1 -16*2.1^2 = 30.24
30.24-32 = -1.76
-1.76/.1 =-17.6
etc getting closer and closer to
dy/dt = v at t = 2 for part b
cheating for that
dy/dt = 48 - 32 t
at t = 2
dy/dt = v = 48 - 64 = -16 :)
@Scott, sketchy lol, Its been since 2018 and I wonder if he's still search for Nick at 7:10
(40(2.5) − 16(2.5)2 − 16)/.5
(a) Ah, it's time for some velocity fun with the good ol' ball in the air! Let's calculate those average velocities one by one.
(i) For the time period of 0.5 seconds, the average velocity is... *drumroll please*... throwing confetti everywhere... 36 ft/s!
(ii) Moving on to a shorter time period of 0.1 seconds, the average velocity is... wait for it... 38.4 ft/s! It's speeding up, folks!
(iii) Now, for an even shorter time period of 0.05 seconds, the average velocity is... beep beep... 39.2 ft/s! It's still accelerating!
(iv) Finally, for a minuscule time period of 0.01 seconds, the average velocity is... ta-da... 39.36 ft/s! It's really picking up speed now!
(b) Alright, let's estimate the instantaneous velocity at t = 2 for our ball in the air act. *puts on a magician's hat* Abracadabra! The instantaneous velocity is approximately 32 ft/s. Now, you see it, now you don't!
To find the average velocity for a given time period, we need to find the displacement between the initial and final positions of the ball and divide it by the time interval.
(a) Average velocity when t = 2 for each time period:
(i) Time interval = 0.5 seconds
To find the average velocity over this time interval, substitute t = 2 and t = 2.5 into the function y = 48t - 16t^2:
y(2.5) = 48(2.5) - 16(2.5)^2 = 120 - 100 = 20 ft
Average velocity = (displacement) / (time interval) = (20 - 0) / (0.5) = 40 ft/s
(ii) Time interval = 0.1 seconds
Substitute t = 2 and t = 2.1 into the function y = 48t - 16t^2:
y(2.1) = 48(2.1) - 16(2.1)^2 = 100.8 - 72.24 = 28.56 ft
Average velocity = (displacement) / (time interval) = (28.56 - 0) / (0.1) = 285.6 ft/s
(iii) Time interval = 0.05 seconds
Substitute t = 2 and t = 2.05 into the function y = 48t - 16t^2:
y(2.05) = 48(2.05) - 16(2.05)^2 = 97.2 - 66.62 = 30.58 ft
Average velocity = (displacement) / (time interval) = (30.58 - 0) / (0.05) = 611.6 ft/s
(iv) Time interval = 0.01 seconds
Substitute t = 2 and t = 2.01 into the function y = 48t - 16t^2:
y(2.01) = 48(2.01) - 16(2.01)^2 = 96.48 - 64.5796 = 31.9004 ft
Average velocity = (displacement) / (time interval) = (31.9004 - 0) / (0.01) = 3190.04 ft/s, rounded to two decimal places.
(b) To estimate the instantaneous velocity when t = 2, we can take the derivative of the position function y with respect to t, which gives us the velocity function v(t).
dy/dt = v(t) = 48 - 32t
Evaluate the velocity function v(t) at t = 2:
v(2) = 48 - 32(2) = 48 - 64 = -16 ft/s
Therefore, the estimated instantaneous velocity at t = 2 is -16 ft/s.