Explanation Given: h ( t ) = 18 t − 9.8 t 2 and u=18 m/s where u is the initial velocity. Formula used: For part a: Average height = 1 t 2 − t 1 ∫ t 1 t 2 h ( t ) d t over the time period [ t 1 , t 2 ] For part b: Average speed= Total distance covered/ total time taken Calculation: By using the equation of motion we find the time interval v = u + a t (where v is the final velocity , u is the initial velocity , a is an acceleration and t is time) Here the final velocity will be 0 because the ball will be at rest at its maximum height 0 = 18 + ( − 9.8 ) × t t = 18 9.8 =1.83 s ……………(1) Now we get the time interval [ t 1 , t 2 ]=[0, 1.83] For part a: Average height = 1 t 2 − t 1 ∫ t 1 t 2 h ( t ) d t = 1 1.83 − 0 ∫ 0 1.83 ( 18 t − 9.8 t 2 ) d t = 1 1.83 ∫ 0 1.83 ( 18 t − 9.8 t 2 ) d t = 1 1.83 ∫ 0 1.83 18 t d t − 1 1.83 ∫ 0 1.83 9.8 t 2 d t = 1 1.83 × 18 ∫ 0 1.83 t d t − 1 1...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 6.2, Problem 54E

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To find: The average height and average speed over the time interval from the ball release to its return to ground level thrown in the air vertically from ground level with initial velocity 18 m/s has height $h (t) = 18 t - 9.8 t^{2}$ at time t.

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Chapter 6.2, Problem 53E

Chapter 6.2, Problem 55E

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You car is moving along in cruise control at a constant speed of 88.3 ft/s (about 60 mi/hr). A text message appears on your phone and you look down to read it. When you look back up at the road, you see a car at a complete stop in your lane 156 ft ahead of you. It takes 0.39 s for you to react and apply the brakes. When the brakes are applied your car experiences an acceleration of -32.3 ft/s2. Fortunately, you are able to avoid hitting the other car. What is the distance between the two cars when you come to a complete stop? (0.9 ft).

Chapter 6 Solutions

Calculus - Standalone book

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The average height and average speed over the time interval from the ball release to its return to ground level thrown in the air vertically from ground level with initial velocity 18 m/s has height h ( t ) = 18 t − 9.8 t 2 at time t.

Solution Summary: The author calculates the average height and speed over the time interval from the ball release to its return to ground level with initial velocity 18 m/s.

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To find: The average height and average speed over the time interval from the ball release to its return to ground level thrown in the air vertically from ground level with initial velocity 18 m/s has height $h (t) = 18 t - 9.8 t^{2}$ at time t.

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Chapter 6 Solutions

The average height and average speed over the time interval from the ball release to its return to ground level thrown in the air vertically from ground level with initial velocity 18 m/s has height h ( t ) = 18 t − 9.8 t 2 at time t.

Solution Summary: The author calculates the average height and speed over the time interval from the ball release to its return to ground level with initial velocity 18 m/s.

Concept explainers

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To find: The average height and average speed over the time interval from the ball release to its return to ground level thrown in the air vertically from ground level with initial velocity 18 m/s has height h(t)=18t−9.8t2 at time t.

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Chapter 6 Solutions

To find: The average height and average speed over the time interval from the ball release to its return to ground level thrown in the air vertically from ground level with initial velocity 18 m/s has height $h (t) = 18 t - 9.8 t^{2}$ at time t.