Textbook Question
Acceleration A drag racer accelerates at a(t)=88 ft/s². Assume v(0)=0, s(0)=0, and t is measured in seconds.
e. How far has the racer traveled when it reaches a speed of 178 ft/s?
Ch. 6 - Applications of Integration
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 6, Problem 6.1.37d
Acceleration A drag racer accelerates at a(t)=88 ft/s². Assume v(0)=0, s(0)=0, and t is measured in seconds.
d. How long does it take the racer to travel 300 ft?
Verified step by step guidance1
Identify the given acceleration function: \(a(t) = 88\) ft/s², which is constant.
Since acceleration is the derivative of velocity, integrate \(a(t)\) with respect to \(t\) to find the velocity function: \(v(t) = \int a(t) \, dt = \int 88 \, dt\).
Use the initial condition \(v(0) = 0\) to solve for the constant of integration in the velocity function.
Next, integrate the velocity function \(v(t)\) with respect to \(t\) to find the position function \(s(t)\): \(s(t) = \int v(t) \, dt\).
Use the initial condition \(s(0) = 0\) to solve for the constant of integration in the position function, then set \(s(t) = 300\) ft and solve for \(t\) to find the time it takes to travel 300 ft.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Acceleration and Its Relationship to Velocity and Position
Acceleration is the rate of change of velocity with respect to time. Given acceleration a(t), velocity v(t) can be found by integrating a(t). Similarly, position s(t) is found by integrating velocity. Understanding these relationships allows us to move from acceleration to position over time.
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Derivatives Applied To Acceleration
Initial Conditions in Integration
Initial conditions such as v(0) = 0 and s(0) = 0 provide specific values needed to solve the constants of integration when finding velocity and position functions. These conditions ensure the solution matches the physical scenario described.
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Solving for Time from Position Function
Once the position function s(t) is determined, solving for the time t when s(t) equals a given distance (300 ft) involves algebraic manipulation. This step finds the exact time required for the racer to travel the specified distance.
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Related Practice
Textbook Question
Bike race Theo and Sasha start at the same place on a straight road, riding bikes with the following velocities (measured in mi/hr). Assume t is measured in hours.
Theo: vT(t)=10, for t≥0
Sasha: vS(t)=15t, for 0≤t≤1, and vS(t)=15, for t>1
f. Suppose Sasha gives Theo a head start of 0.2 hr and the riders ride for 20 mi. Who wins the race?
Textbook Question
Displacement and distance from velocity Consider the velocity function shown below of an object moving along a line. Assume time is measured in seconds and distance is measured in meters. The areas of four regions bounded by the velocity curve and the t-axis are also given.
e. Describe the position of the object relative to its initial position after 8 seconds.
Textbook Question
9–10. Velocity graphs The figures show velocity functions for motion along a line. Assume the motion begins with an initial position of s(0)=0. Determine the following.
d. A piecewise function for s(t)
Textbook Question
Displacement and distance from velocity Consider the velocity function shown below of an object moving along a line. Assume time is measured in seconds and distance is measured in meters. The areas of four regions bounded by the velocity curve and the t-axis are also given.
d. What is the displacement of the object over the interval [0, 8]?
Textbook Question
Determine whether the following statements are true and give an explanation or counterexample.
d. Let f(x)=12x^2.. The area of the surface generated when the graph of f on [−4, 4] is revolved about the y-axis is twice the area of the surface generated when the graph of f on [0, 4] is revolved about the y-axis.