The distance traveled by the ball as a function of time, s(t) = 16•t² ft., and the rate of change, based on calculus, gives;
(a) During the time interval [6, 6.5], the distance traveled is 100 feet
(b) In the interval, [6, 6.5], the average velocity is 200 ft/s
(c) The calculated average velocities in ft/s, for the respective time intervals are;
[6, 6.01]. 192.16
[6, 6.001]. 192.016
[6, 6.0001]. 192.0016
[5.9999, 6]. 191.9984
[5.999, 6]. 191.984
[5.99, 6]. 191.84
The instantaneous velocity at t = 6 is v(6) = 192 ft/s
What are the average and instantaneous velocities of the ball?
The distance traveled as a function of time is s(t) = 16•t² ft.
(a) The distance traveled during the time interval [6, 6.5] is found as follows;
s(6) = 16 × 6² = 576
s(6.5) = 16 × 6.5² = 676
The distance traveled in the interval, [6, 6.5], ∆s(t) = 676 - 576 = 100
The distance traveled during the time interval [6, 6.5] = 100 feet
(b) Average velocity = ∆s(t)/∆t
∆t = 6.5 - 6 = 0.5
Therefore;
∆s(t)/∆t = 100/0.5 = 200
The average velocity = 200 ft./s
(c) The average velocities over the given time intervals, obtained using a graphing calculator, are;
Interval. Average velocity
[6, 6.01]. 192.16
[6, 6.001]. 192.016
[6, 6.0001]. 192.0016
[5.9999, 6]. 191.9984
[5.999, 6]. 191.984
[5.99, 6]. 191.84
Given that at t = 6, as the time interval approaches zero from the right and from the left, the average velocity approaches 192 ft./s, the instantaneous velocity at t = 6, v(6) is 192 ft./s
v(6) = 192 ft./s
Using calculus, differentiation, we have;
v(t) = ds/dt = 2×16 × t = 32•t
v(6) = 32 × 6 = 192
Therefore;
v(6) = 192 ft./s
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