The position of an object moving vertically along a line is given by the function s(t)= −16t^2 +128t. Find the average velocity of the object over the following intervals.
(A) [1, 4],
(B) [1, 3],
(C) [1, 2],
(D) [1, 1+h], where h<0 is a real number.

a) To find the distance it moved, we need to know the position the object was at the start of the interval and the position it was at the end of the interval. I will suppose the position is in meters and time is in seconds.

The position at the start of the interval is s(1) = -16*1² + 128(1) = 112m

The position at the end of the interval is s(4) = -16*4² + 128(4) = 256m

So, from the start to the end of the interval, the object moved 256-112 = 144m in 4-1 = 3 seconds.

Av = 144/3 = 48 m/s

b) From a), we already know s(1) = 112.

We need to find s(3)

s(3) = -16*3² + 128*3 = -144 + 384 = 240m.

From the start to the end of the interval, the object moved 240-112 = 128m in 3-1 = 2 seconds.

Av = 128/2 = 64 m/s

c) From a), s(1) = 112

We need to find s(2)

s(2) = -16*2² + 128*2 = -64 + 256 = 192m

From the start to the end of the interval, the object moved 192-112 = 80m in 2-1 = 1 second.

Av = 80/1 = 80m/s

d) From a), s(1) = 112

We need to find s(1+h).

s(1+h) = -16*(1+h)² + 128(1+h) = -16(1 + 2h + h²) + 128 + 128h = -16 - 32h - 16h² + 128 + 128h = -16h² + 96h + 112.

From the start to the end of the interval, the object moved -16h² + 96h + 112 - 112 = (-16h² + 96h)m in 1+h-1 = h seconds.

Av = (-16h² + 96h)/h = h(-16h+96)/h = -16h + 96

The position of an object moving vertically along a line is given by the function s(t)= −16t^2 +128t. Find the average velocity of the object over the following intervals. (A) [1, 4], (B) [1, 3], (C) [1, 2], (D) [1, 1+h], where h<0 is a real number.

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The position of an object moving vertically along a line is given by the function s(t)= −16t^2 +128t. Find the average velocity of the object over the following intervals.
(A) [1, 4],
(B) [1, 3],
(C) [1, 2],
(D) [1, 1+h], where h<0 is a real number.