Chapter 4 | Applications of Derivatives
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given by the function s ( t ) =−16 t 2 +100. a. Determine how long it takes before the rock hits the ground.
b. Find the average velocity v avg of the rock for when the rock is released and the rock hits the ground. c. Find the time t guaranteed by the Mean Value Theorem when the instantaneous velocity of the rock is v avg . Solution a. When the rock hits the ground, its position is s ( t ) =0. Solving the equation −16 t 2 +100=0 for t , we find that t =± 5 2 sec. Since we are only considering t ≥0, the ball will hit the ground 5 2 sec after it is dropped. b. The average velocity is given by v avg = s (5/2)− s (0) 5/2−0 = 0−100 5/2 = −40 ft/sec. c. The instantaneous velocity is given by the derivative of the position function. Therefore, we need to find a time t such that v ( t ) = s ′( t ) = v avg = −40 ft/sec. Since s ( t ) is continuous over the interval [0, 5/2] and differentiable over the interval (0, 5/2), by the Mean Value Theorem, there is guaranteed to be a point c ∈ (0, 5/2) such that s ′( c ) = s (5/2)− s (0) 5/2−0 =−40. Taking the derivative of the position function s ( t ), we find that s ′( t ) =−32 t . Therefore, the equation reduces to s ′( c ) =−32 c =−40. Solving this equation for c , we have c = 5 4 . Therefore, 5 4 sec after the rock is dropped, the instantaneous velocity equals the average velocity of the rock during its free fall: −40 ft/sec.
Figure 4.28 At time t =5/4 sec, the velocity of the rock is equal to its average velocity from the time it is dropped until it hits the ground.
4.15 Suppose a ball is dropped from a height of 200 ft. Its position at time t is s ( t ) =−16 t 2 +200. Find the time t when the instantaneous velocity of the ball equals its average velocity.
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