/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 The motion of a particle is defi... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 6

The motion of a particle is defined by the relation \(x=t^{3}-9 t^{2}+\) \(24 t-8,\) where \(x\) and \(t\) are expressed in inches and seconds, respectively. Determine (a) when the velocity is zero, \((b)\) the position and the total distance traveled when the acceleration is zero.

Short Answer

Expert verified
Velocity is zero at \(t = 2\) and \(t = 4\). Position is 10 inches, total distance is 16 inches.

Step by step solution

01

Find the velocity equation

The velocity is the first derivative of the position function. Differentiate the position equation \(x = t^{3} - 9t^{2} + 24t - 8\) with respect to \(t\). This gives \(v(t) = \frac{d}{dt}(t^3 - 9t^2 + 24t - 8) = 3t^2 - 18t + 24\).
02

Determine when velocity is zero

Set the velocity equation \(v(t) = 3t^2 - 18t + 24 = 0\) and solve for \(t\). Factor the quadratic to get \(3(t^2 - 6t + 8) = 0\), which can be factored further as \(3(t-2)(t-4) = 0\). Thus, the velocity is zero at \(t = 2\) and \(t = 4\) seconds.
03

Calculate the acceleration equation

Acceleration is the derivative of velocity. Differentiate the velocity equation \(v(t) = 3t^2 - 18t + 24\) to find the acceleration: \(a(t) = \frac{d}{dt}(3t^2 - 18t + 24) = 6t - 18\).
04

Determine when acceleration is zero

Set the acceleration equation \(a(t) = 6t - 18 = 0\) and solve for \(t\). Solve \(6t = 18\) to get \(t = 3\) seconds.
05

Find position when acceleration is zero

Substitute \(t = 3\) back into the position equation: \(x = 3^3 - 9(3)^2 + 24(3) - 8 = 27 - 81 + 72 - 8\). Simplifying, \(x = 10\) inches.
06

Calculate total distance traveled between \(t = 0\) and \(t = 4\)

Analyze the intervals where changes occur鈥攙elocity changes sign at \(t = 2\). Compute distances: From \(t = 0\) to \(t = 2\), find \(x(0) = -8\) and \(x(2) = 0\); distance is \(|0 - (-8)| = 8\) inches. From \(t = 2\) to \(t = 4\), find \(x(4) = 8\); the distance is \(|8 - 0| = 8\) inches. Total distance = 8 + 8 = 16 inches.

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Key Concepts

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

Velocity Equation in Particle Motion
Understanding the velocity equation is crucial when studying particle motion. Velocity describes how fast the position of a particle changes over time. It is essentially the rate of change of displacement. To find the velocity, you need to take the derivative of the position function.For example, if the position of a particle is given by the function \(x = t^3 - 9t^2 + 24t - 8\), the velocity equation \(v(t)\) is obtained by differentiating this equation with respect to time \(t\). This results in \(v(t) = 3t^2 - 18t + 24\).By solving the velocity equation, particularly by setting it equal to zero, you can find the moments in time (\(t\)) when the particle's velocity is zero, indicating a change in the direction of motion. In our example, the velocity is zero at \(t = 2\) and \(t = 4\) seconds.
Understanding When Acceleration is Zero
Acceleration measures the rate of change of velocity. In other words, it tells us how quickly a particle is picking up speed or slowing down. When acceleration is zero, the particle moves at a constant velocity.To find out when this happens, take the derivative of the velocity function. From our example, differentiating \(v(t) = 3t^2 - 18t + 24\) gives the acceleration equation \(a(t) = 6t - 18\). Set \(a(t) = 0\) to find when the acceleration is zero.Upon solving \(6t - 18 = 0\), you get \(t = 3\) seconds. This means at \(t = 3\) seconds, the acceleration is zero, and the particle is moving with a constant speed. Substituting \(t = 3\) into the original position function \(x = t^3 - 9t^2 + 24t - 8\), you find that the particle is at \(x = 10\) inches at this moment.
Calculating Total Distance Traveled
The total distance traveled by a particle refers to the absolute total of how much ground it covers over a period of time. It's important to note that this differs from displacement, which only concerns the change in position from start to finish.To calculate the total distance, identify intervals where the particle's velocity changes sign, as this indicates a change in direction. In our example, the velocity is zero at \(t = 2\) and \(t = 4\) seconds, suggesting two motion intervals: from \(t = 0\) to \(t = 2\) and from \(t = 2\) to \(t = 4\).
  • First, calculate the position at these critical points: \(x(0) = -8\), \(x(2) = 0\), and \(x(4) = 8\) inches.
  • Next, determine the distance for each interval:
  • The distance from \(t = 0\) to \(t = 2\) is \(|0 - (-8)| = 8\) inches.
  • The distance from \(t = 2\) to \(t = 4\) is \(|8 - 0| = 8\) inches.
Adding these distances gives you a total distance of 16 inches. This shows that while the particle's displacement might differ, the total distance tells you the complete journey.

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Most popular questions from this chapter

An airplane begins its take-off run at \(A\) with zero velocity and a constant acceleration \(a\). Knowing that it becomes airborne 30 s later at \(B\) and that the distance \(A B\) is \(900 \mathrm{m}\), determine ( \(a\) ) the acceleration \(a\) (b) the take-off velocity \(v_{B}\).

A golfer hits a golf ball with an initial velocity of \(160 \mathrm{ft} / \mathrm{s}\) at an angle of \(25^{\circ}\) with the horizontal. Knowing that the fairway slopes downward at an average angle of \(5^{\circ}\), determine the distance \(d\) between the golfer and point \(B\) where the ball first lands.

Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit \(100 \mathrm{mi}\) above the surface of the planet. (See information given in Probs. \(11.153-11.154 .)\) $$ \begin{array}{l}{11.155 \text { Venus: } g=29.20 \mathrm{ft} / \mathrm{s}^{2}, R=3761 \mathrm{mi}} \\ {11.156 \mathrm{Mars}: g=12.17 \mathrm{ft} / \mathrm{s}^{2}, R=2102 \mathrm{mi}} \\ {11.157 \text { Jupiter: } g=75.35 \mathrm{ft} / \mathrm{s}^{2}, R=44,432 \mathrm{mi}}\end{array} $$

Instruments in airplane \(A\) indicate that, with respect to the air, the plane is headed \(30^{\circ}\) north of east with an air speed of \(300 \mathrm{mi} / \mathrm{h}\). At the same time, radar on ship \(B\) indicates that the relative velocity of the plane with respect to the ship is \(280 \mathrm{mi} / \mathrm{h}\) in the direction \(33^{\circ}\) north of east. Knowing that the ship is steaming due south at \(12 \mathrm{mi} / \mathrm{h}\), determine ( \(a\) ) the velocity of the airplane, \((b)\) the wind speed and direction.

Car \(A\) is traveling at 40 milh when it enters a 30 milh speed zone. The driver of car \(A\) decelerates at a rate of \(16 \mathrm{ft} / \mathrm{s}^{2}\) until reaching a speed of \(30 \mathrm{mil}\), which she then maintains. When car \(B\), which was initially \(60 \mathrm{ft}\) behind car \(A\) and traveling at a constant speed of \(45 \mathrm{mil}\), enters the speed zone, its driver decelerates at a rate of \(20 \mathrm{ft}\) s " until reaching a speed of \(28 \mathrm{mi} / \mathrm{h}\). Knowing that the driver of car \(B\) maintains a speed of \(28 \mathrm{mi} / \mathrm{h}\), determine \((a)\) the closest that car \(B\) comes to car \(A,(b)\) the time at which car \(A\) is \(70 \mathrm{ft}\) in front of car \(B .\)
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