The average velocity of an object moving in a straight line is the total displacement (the change in position) divided by the total time taken. In this case, the displacement is given by the difference in the height function y = 50t - 16t^2 at the two time points.

(a)
(i) For the time period beginning when t = 2 and lasting 0.5 second, the average velocity is:

(y(2.5) - y(2)) / (2.5 - 2) = (50*2.5 - 16*2.5^2 - (50*2 - 16*2^2)) / 0.5 = -24 ft/s

(ii) For the time period beginning when t = 2 and lasting 0.1 second, the average velocity is:

(y(2.1) - y(2)) / (2.1 - 2) = (50*2.1 - 16*2.1^2 - (50*2 - 16*2^2)) / 0.1 = -28.4 ft/s

(iii) For the time period beginning when t = 2 and lasting 0.05 second, the average velocity is:

(y(2.05) - y(2)) / (2.05 - 2) = (50*2.05 - 16*2.05^2 - (50*2 - 16*2^2)) / 0.05 = -29.2 ft/s

(iv) For the time period beginning when t = 2 and lasting 0.01 second, the average velocity is:

(y(2.01) - y(2)) / (2.01 - 2) = (50*2.01 - 16*2.01^2 - (50*2 - 16*2^2)) / 0.01 = -29.84 ft/s

(b) The instantaneous velocity when t = 2 can be estimated as the limit of the average velocity as the time interval approaches 0. From the calculations above, we can see that this limit is approximately -30 ft/s. For a more precise value, we can use the derivative of the height function, which gives the instantaneous velocity as a function of time. The derivative of y = 50t - 16t^2 is y' = 50 - 32t, so the instantaneous velocity when t = 2 is y'(2) = 50 - 32*2 = -14 ft/s.

Question

The average velocity of an object moving in a straight line is the total displacement (the change in position) divided by the total time taken. In this case, the displacement is given by the difference in the height function y = 50t - 16t^2 at the two time points.

(a) 
(i) For the time period beginning when t = 2 and lasting 0.5 second, the average velocity is:

(y(2.5) - y(2)) / (2.5 - 2) = (50*2.5 - 16*2.5^2 - (50*2 - 16*2^2)) / 0.5 = -24 ft/s

(ii) For the time period beginning when t = 2 and lasting 0.1 second, the average velocity is:

(y(2.1) - y(2)) / (2.1 - 2) = (50*2.1 - 16*2.1^2 - (50*2 - 16*2^2)) / 0.1 = -28.4 ft/s

(iii) For the time period beginning when t = 2 and lasting 0.05 second, the average velocity is:

(y(2.05) - y(2)) / (2.05 - 2) = (50*2.05 - 16*2.05^2 - (50*2 - 16*2^2)) / 0.05 = -29.2 ft/s

(iv) For the time period beginning when t = 2 and lasting 0.01 second, the average velocity is:

(y(2.01) - y(2)) / (2.01 - 2) = (50*2.01 - 16*2.01^2 - (50*2 - 16*2^2)) / 0.01 = -29.84 ft/s

(b) The instantaneous velocity when t = 2 can be estimated as the limit of the average velocity as the time interval approaches 0. From the calculations above, we can see that this limit is approximately -30 ft/s. For a more precise value, we can use the derivative of the height function, which gives the instantaneous velocity as a function of time. The derivative of y = 50t - 16t^2 is y' = 50 - 32t, so the instantaneous velocity when t = 2 is y'(2) = 50 - 32*2 = -14 ft/s.

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