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Chapter 6: Problem 12

Suppose an object moves along a line with velocity (in \(\mathrm{ft} / \mathrm{s}\) ) \(v(t)=6-2 t,\) for \(0 \leq t \leq 6\) where \(t\) is measured in seconds. a. Graph the velocity function on the interval \(0 \leq t \leq 6\) Determine when the motion is in the positive direction and when it is in the negative direction on \(0 \leq t \leq 6\). b. Find the displacement of the object on the interval \(0 \leq t \leq 6\). c. Find the distance traveled by the object on the interval \(0 \leq t \leq 6\).

Short Answer

Expert verified
Answer: The motion is positive in the interval \(0 \leq t \leq 3\), and negative in the interval \(3 \leq t \leq 6\). The displacement of the object is 0 feet, and the distance traveled by the object is 6 feet.

Step by step solution

01

Graph the velocity function and find critical points

We are given the velocity function as \(v(t) = 6 - 2t\). Let's graph this function and find the critical points. To find the critical points, we need to solve \(v(t) = 0\): $$6 - 2t = 0$$ $$t = 3$$ Now, we have one critical point at \(t = 3\). This divides the interval \(0 \leq t \leq 6\) into two subintervals: \(0 \leq t \leq 3\) and \(3 \leq t \leq 6.\)
02

Determine the motion direction in the intervals

We will now analyze the behavior of the velocity function in the subintervals obtained in the previous step: - For \(0 \leq t \leq 3\): We can choose any point in this interval, say \(t = 1\), and evaluate the velocity function: \(v(1) = 6 - 2(1) = 4\), which is positive. So, the motion is in the positive direction. - For \(3 \leq t \leq 6\): We can choose any point in this interval, say \(t = 4\), and evaluate the velocity function: \(v(4) = 6 - 2(4) = -2\), which is negative. Therefore, the motion is in the negative direction.
03

Find the displacement

To find the displacement, we need to integrate the velocity function over the interval \(0 \leq t \leq 6\): $$\text{Displacement} = \int_0^6 v(t) dt = \int_0^6 (6 - 2t) dt$$ Using the integration rules, we can obtain: $$\text{Displacement} = \left[6t - t^2\right]_{0}^{6} = (6(6) - (6)^2 ) - (6(0) - (0)^2) = 36 - 36 = 0$$ The displacement of the object over the interval \(0 \leq t \leq 6\) is 0 feet.
04

Find the distance traveled

To find the distance traveled, we need to integrate the absolute value of the velocity function over the interval \(0 \leq t \leq 6\). Since we have already found that the motion is in the positive direction when \(0 \leq t \leq 3\) and negative when \(3 \leq t \leq 6\), we can split the integral into two parts: $$\text{Distance Traveled} = \int_0^3 4 -2t dt - \int_3^6 6-2t dt$$ Now, we integrate each part: $$\int_0^3 4-2t dt = \left[4t - t^2\right]_{0}^{3} = 12 - 9 =3$$ $$-\int_3^6 6-2t dt = -[\left(6t - t^2\right]_{3}^{6}) =3$$ Here, we used a negative sign in the second integral to compensate the negative direction of the motion. Now, we sum both parts: $$\text{Distance Traveled} = 3 + 3 = 6$$ The distance traveled by the object over the interval \(0 \leq t \leq 6\) is 6 feet.

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Key Concepts

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

Velocity Function
In calculus, the velocity function represents how an object's velocity changes over time. For our specific problem, we have the velocity function given as \( v(t) = 6 - 2t \). The notation \( v(t) \) indicates that velocity is dependent on time \( t \). This function is linear, which means it describes a straight line when graphed.
An essential step in understanding such problems is to determine the critical points where velocity changes direction. This is found by setting the velocity equation to zero (\( v(t) = 0 \)) and solving for \( t \). Here, we solve \( 6 - 2t = 0 \), leading to \( t = 3 \).
This critical point divides the time interval \( 0 \leq t \leq 6 \) into two sections:
  • \( 0 \leq t < 3 \) - Velocity is positive, causing forward motion.
  • \( 3 \leq t \leq 6 \) - Velocity is negative, resulting in backward motion.
This change in direction is crucial in calculating both displacement and distance traveled.
Displacement
Displacement refers to the object's overall change in position. Unlike distance, displacement considers the direction. We calculate it by integrating the velocity over the given time interval.
In this problem, to find displacement over \( 0 \leq t \leq 6 \), we compute the integral \( \int_0^6 (6 - 2t) \, dt \). This results in \( 6t - t^2 \) evaluated from 0 to 6, which equals zero.

The zero displacement indicates that the object finishes at its starting point, effectively canceling out any forward or backward movements. This result does not mean the object did not move; rather, it ended up in the same position where it began.
This illustrates the concept that displacement, being a vector quantity, highlights the net change in position, not the distance covered.
Distance Traveled
The distance traveled by an object accounts for all the movement regardless of direction, making it a scalar quantity. Calculating distance entails considering the absolute value of the velocity function.
Since the problem involves forward and backward motion, we separate the calculation into parts:
  • For \( 0 \leq t < 3 \), the velocity is positive, so distance is \( \int_0^3 (6 - 2t) \, dt \).
  • For \( 3 \leq t \leq 6 \), velocity is negative. The distance becomes \( -\int_3^6 (6 - 2t) \, dt \) to ensure it's positive.

The integrals calculate to 3 feet each, contributing to a total distance of 6 feet traveled by the object.
This thorough separation and calculation ensure that every part of the object's journey is accounted for, demonstrating distance as independent of initial or final position.

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Most popular questions from this chapter

Suppose \(f\) is a nonnegative function with a continuous first derivative on \([a, b] .\) Let \(L\) equal the length of the graph of \(f\) on \([a, b]\) and let \(S\) be the area of the surface generated by revolving the graph of \(f\) on \([a, b]\) about the \(x\) -axis. For a positive constant \(C\), assume the curve \(y=f(x)+C\) is revolved about the \(x\) -axis. Show that the area of the resulting surface equals the sum of \(S\) and the surface area of a right circular cylinder of radius \(C\) and height \(L\) 1\. The surface area of the first cone \((200 \sqrt{5} \pi)\) is twice as great as the surface area of the second cone \((100 \sqrt{5} \pi)\) 2\. The surface area is \(63 \sqrt{10} \pi .\) 3. The surface is a cylinder of radius \(c\) and height \(b-a .\) The area of the curved surface is \(2 \pi c(b-a)\)

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