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Chapter 6: Problem 3

Given the velocity function v of an object moving along a line. explain how definite integrals can be used to find the displacement of the object.

Short Answer

Expert verified
Answer: The displacement of an object moving along a line can be found by calculating the definite integral of the given velocity function v(t) with respect to time 't' over the interval [a, b], where 'a' represents the initial time and 'b' represents the final time. Displacement is the total change in position of the object, and since the velocity function represents the rate of change of position, integrating the velocity function over the given time interval yields the displacement as follows: Displacement = 鈭玔a, b] v(t) dt = V(b) - V(a).

Step by step solution

01

Define Displacement

Displacement is the change in position of an object over a period of time. To find the displacement of an object, we need to find the difference between its initial position (x_initial) and its final position (x_final). Mathematically, Displacement can be defined as: Displacement = x_final - x_initial.
02

Understand the velocity function

The given velocity function, v(t), models the rate of change of the position of the object with respect to time. Simply put, the velocity function tells us how fast the position of the object is changing at any given time 't'.
03

Link velocity to displacement

Since the velocity function represents the rate of change of position over time, the displacement of the object can be found by integrating the velocity function with respect to time over a given interval [a, b], where 'a' represents the initial time, and 'b' represents the final time. Mathematically, Displacement can be expressed as: Displacement = 鈭玔a, b] v(t) dt.
04

Calculate the definite integral

To find the displacement of the object over the given interval [a, b], the definite integral of the velocity function, v(t), with respect to 't' must be calculated. Thus, compute the integral by finding the antiderivative of the velocity function V(t) and evaluating it at the limits 'a' and 'b': Displacement = V(b) - V(a).
05

Conclusion

In summary, the displacement of the object over a given time interval [a, b], can be found by calculating the definite integral of the given velocity function v(t) with respect to time 't' over the interval [a, b]. This is because the velocity function represents the rate of change of position of the object, and taking the definite integral of the velocity function over the interval [a, b] gives the total change in position or the displacement of the object.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Definite Integral
The definite integral is a fundamental concept in calculus that helps calculate the total accumulation of a quantity. In simpler terms, it's like adding up small pieces to find a whole.
When you integrate a function over a specific interval, you are essentially summing up all the tiny changes described by the function over that period.
  • Interval Concept: The interval \[a, b\] defines the start and end points on the time axis over which you're interested in accumulating the changes.
  • Mathematical Notation: The definite integral is written as \int_{a}^{b} f(x) \, dx\ which reads as "the integral of function \( f(x) \) from \( a \) to \( b \)."
  • Application: In the context of displacement, it accumulates changes in velocity to determine the net change in position.
The definite integral provides a precise measurement of the accumulation, making it critical for solving physical problems, such as finding displacement from velocity.
Exploring the Velocity Function
The velocity function, usually denoted as \( v(t) \), plays a crucial role in motion analysis. It describes how fast an object's position is changing over time.
This function gives insights into the object's speed and direction at any given moment.
  • Rate of Change: Velocity is the rate of change of an object's position with respect to time, providing a dynamic view of motion.
  • Significance in Displacement: Using the velocity function, you can determine how far an object has moved by integrating it over a specific time interval.
  • Practical Example: If a car's velocity is described by \( v(t) = 3t^2 \), we can find out how far it has traveled in a specific timeframe by integrating \( v(t) \).
The velocity function is essential as it is the starting point for calculating displacement through its integration.
Antiderivative and Its Role
The antiderivative, also known as the indefinite integral, is the reverse process of differentiation. It is an essential component because it allows us to find a function whose derivative gives the velocity function.
This allows us to connect velocity back to position.
  • Finding the Antiderivative: To find the displacement, you first find the antiderivative \( V(t) \) of the velocity function \( v(t) \).
  • Definite Integral Connection: By evaluating the antiderivative at the limits of integration \( a \) and \( b \), you find the total change in position: \( V(b) - V(a) \).
  • Simplifying Calculations: The antiderivative simplifies calculating the definite integral, providing the accumulation needed for displacement.
The antiderivative is like a bridge, linking the rate of change (velocity) back to the original function (position), making it a pivotal tool in calculus.

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Most popular questions from this chapter

Suppose \(f\) and \(g\) have continuous derivatives on an interval \([a, b] .\) Prove that if \(f(a)=g(a)\) and \(f(b)=g(b),\) then \(\int_{a}^{b} f^{\prime}(x) d x=\int_{a}^{b} g^{\prime}(x) d x\).

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The mass of a thin wire is the length of the wire times its average density over its length. b. The work required to stretch a linear spring (that obeys Hooke's law) \(100 \mathrm{cm}\) from equilibrium is the same as the work required to compress it \(100 \mathrm{cm}\) from equilibrium. c. The work required to lift a 10 -kg object vertically \(10 \mathrm{m}\) is the same as the work required to lift a 20 -kg object vertically 5 m. d. The total force on a \(10-f t^{2}\) region on the (horizontal) floor of a pool is the same as the total force on a 10 -ft \(^{2}\) region on a (vertical) wall of the pool.

A large tank has a plastic window on one wall that is designed to withstand a force of 90,000 N. The square window is 2 m on a side, and its lower edge is 1 m from the bottom of the tank. a. If the tank is filled to a depth of 4 m, will the window with-stand the resulting force? b. What is the maximum depth to which the tank can be filled without the window failing?

Consider the following velocity functions. In each case, complete the sentence: The same distance could have been traveled over the given time period at a constant velocity of _______. $$v(t)=2 \sin t,\( for \)0 \leq t \leq \pi$$

Where do they meet? Kelly started at noon \((t=0)\) riding a bike from Niwot to Berthoud, a distance of \(20 \mathrm{km}\), with velocity \(v(t)=\frac{15}{(t+1)^{2}}\) (decreasing because of fatigue). Sandy started at noon \((t=0)\) riding a bike in the opposite direction from Berthoud to Niwot with velocity \(u(t)=\frac{20}{(t+1)^{2}}\) (also decreasing because of fatigue). Assume distance is measured in kilometers and time is measured in hours. Make a graph of Kelly's distance from Niwot as a function of time. b. Make a graph of Sandy's distance from Berthoud as a function of time. c. When do they meet? How far has each person traveled when they meet? d. More generally, if the riders' speeds are \(v(t)=\frac{A}{(t+1)^{2}}\) and \(u(t)=\frac{B}{(t+1)^{2}}\) and the distance between the towns is \(D\) what conditions on \(A, B,\) and \(D\) must be met to ensure that the riders will pass each other? Looking ahead: With the velocity functions given in part (d). make a conjecture about the maximum distance each person can ride (given unlimited time).
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