If a ball is thrown straight up into the air with an initial velocity of 95 ft/s, its height in feet after t seconds is given by
f(t)=95t−16t^2
Find the average velocity for the time period beginning when t=1 and lasting
(i) 0.5 seconds
(ii) 0.1 seconds
(iii) 0.01 seconds
(iii) 0.001 seconds
Based on the above results, guess what the instantaneous velocity of the ball is when t=1.
in each case, the average velocity is ∆y/∆x for the given interval.
So, for [1,1.5],
∆y/∆x = (f(1.5)-f(1))/(1.5-1)
= (106.5-79)/.5
= 55
Do the others similarly
To find the average velocity for a given time period, we need to calculate the change in position (or height) and divide it by the change in time.
(i) For a time period of 0.5 seconds:
We need to find the change in position between t=1 and t=1.5. Substitute these values into the equation f(t) = 95t − 16t^2:
f(1) = 95(1) - 16(1)^2 = 79 ft
f(1.5) = 95(1.5) - 16(1.5)^2 = 112.75 ft
The change in position is Δx = f(1.5) - f(1) = 112.75 - 79 = 33.75 ft
The change in time is Δt = 1.5 - 1 = 0.5 s
So, the average velocity is Vavg = Δx / Δt = 33.75 ft / 0.5 s = 67.5 ft/s.
(ii) For a time period of 0.1 seconds:
We need to find the change in position between t=1 and t=1.1:
f(1) = 79 ft
f(1.1) = 95(1.1) - 16(1.1)^2 ≈ 84.65 ft
The change in position is Δx = f(1.1) - f(1) = 84.65 - 79 = 5.65 ft
The change in time is Δt = 1.1 - 1 = 0.1 s
So, the average velocity is Vavg = Δx / Δt = 5.65 ft / 0.1 s = 56.5 ft/s.
(iii) For a time period of 0.01 seconds:
We need to find the change in position between t=1 and t=1.01:
f(1) = 79 ft
f(1.01) = 95(1.01) - 16(1.01)^2 ≈ 79.605 ft
The change in position is Δx = f(1.01) - f(1) = 79.605 - 79 = 0.605 ft
The change in time is Δt = 1.01 - 1 = 0.01 s
So, the average velocity is Vavg = Δx / Δt = 0.605 ft / 0.01 s = 60.5 ft/s.
(iv) For a time period of 0.001 seconds:
We need to find the change in position between t=1 and t=1.001:
f(1) = 79 ft
f(1.001) = 95(1.001) - 16(1.001)^2 ≈ 79.96079 ft
The change in position is Δx = f(1.001) - f(1) = 79.96079 - 79 = 0.96079 ft
The change in time is Δt = 1.001 - 1 = 0.001 s
So, the average velocity is Vavg = Δx / Δt = 0.96079 ft / 0.001 s = 960.79 ft/s.
Based on the above results, we can observe that as the time period decreases, the average velocity approaches the instantaneous velocity. Therefore, when t=1, we can estimate that the instantaneous velocity of the ball is approximately 960.79 ft/s.