What is the total distance traveled by the ball for 0<=t<=2?

s(t) = -16t² + 32t + 6 shows the position of the ball.

When t = 0: s(0) = -16(0)² + 32(0) + 6 = 6 (ft)
When t = 2: s(2) = -16(2)² + 32(2) + 6 = 6 (ft)

This shown that when 0 ≤ t ≤ 2, the ball firstly going up to the highest point, and then falls down to the original position. We have to find the highest position that the stone can reach.


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Method 1 : The parabola is symmetric about the axis of symmetry

Time taken when the ball is going up
= Time taken when the ball is going down
= (2 s) ×(1/2)
= 1 s

When t = 1: s(1) = -16(1)² + 32(1) + 6 = 22 (ft)

Distance travelled by the ball
= 2 × (22 - 6) ft
= 32 ft


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Method 2 : Differentiation

s(t) = -16t² + 32t + 6
s'(t) = -32t + 32 = -32(t - 1)
s"(t) = -32

When t = 1 :
s(t) = -16(1)² + 32t + 6 = 22
s'(t) = 0
s"t = -32 < 0

Hence, the highest position = 22 ft

Distance travelled by the ball
= 2 × (22 - 6) ft
= 32 ft


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Method 3 : Completing square

s(t) = -16t² + 32t + 6
s(t) = -16(t² - 2t) + 6
s(t) = -16(t² - 2t + 1) + 16 + 6
s(t) = -16(t - 1)² + 22

For all real t, -16(t - 1)² ≤ 0
Hence, s(t) = -16(t - 1)² + 22 ≤ 22
Maximum s(t) = 22 (ft)

Distance travelled by the ball
= 2 × (22 - 6) ft
= 32 ft

You are correct in your assessment of the various values of s(t). The total distance traveled will be 16 feet up (from 6 to 22) and the same number down. Twice 16 feet is 32 feet, not 26.

At t=1, the ball has traveled 16 ft; and at t=2, it has traveled 32 ft.

The question is given in heading "What is the total distance traveled by the ball for 0<=t<=2". According to my new calculations the answer is 32