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Chapter 2: Problem 18

The position of the front bumper of a test car under microprocessor control is given by \(x(t)=2.17 \mathrm{m}+\) \(\left(4.80 \mathrm{m} / \mathrm{s}^{2}\right) t^{2}-\left(0.100 \mathrm{m} / \mathrm{s}^{6}\right) t^{6} .\) (a) Find its position and acceleration at the instants when the car has zero velocity. (b) Draw \(x\) -t, \(v_{x}-t,\) and \(a_{x}-t\) graphs for the motion of the bumper between \(t=0\) and \(t=2.00 \mathrm{s}.\)

Short Answer

Expert verified
Position at zero velocity: 2.17 m and 14.97 m; acceleration: 9.60 m/s虏 and -38.40 m/s虏.

Step by step solution

01

Determine the Velocity Function

The velocity function is the derivative of the position function with respect to time, \( v(t) = \frac{dx(t)}{dt} \). Given \( x(t) = 2.17 + 4.80t^2 - 0.100t^6 \), differentiate to find \( v(t) = 9.60t - 0.600t^5 \).
02

Find Instants of Zero Velocity

Set the velocity function equal to zero and solve for \( t \).\[ 0 = 9.60t - 0.600t^5 \]Factor out \( t \),\[ t(9.60 - 0.600t^4) = 0 \]This gives \( t = 0 \) and solving \( 9.60 - 0.600t^4 = 0 \) gives \( t^4 = 16 \), so \( t = 2 \) sec (since negative \( t \) is not considered in our interval).
03

Calculate Position at Zero Velocity

Using the times from Step 2, substitute them into the position function to get the corresponding positions.\For \( t = 0 \),\[ x(0) = 2.17 \text{ meters} \]For \( t = 2 \),\[ x(2) = 2.17 + 4.80(2)^2 - 0.100(2)^6 = 2.17 + 19.2 - 6.4 = 14.97 \text{ meters} \]
04

Determine the Acceleration Function

The acceleration function is the derivative of the velocity function with respect to time, \( a(t) = \frac{dv(t)}{dt} \). From \( v(t) = 9.60t - 0.600t^5 \),\[ a(t) = 9.60 - 3.00t^4 \].
05

Calculate Acceleration at Zero Velocity

Substitute \( t = 0 \) and \( t = 2 \) into the acceleration function to determine the accelerations.\For \( t = 0 \),\[ a(0) = 9.60 \text{ m/s}^2 \]For \( t = 2 \),\[ a(2) = 9.60 - 3.00(2)^4 = 9.60 - 48 = -38.40 \text{ m/s}^2 \]
06

Sketch the Graphs

Using the functions derived, plot the position-time, velocity-time, and acceleration-time graphs for the interval \( t = 0 \) to \( t = 2.00 \)s.1. The \( x-t \) graph is parabolic, starting at 2.17m and passing through the calculated points.2. The \( v_x-t \) graph starts at zero, reaches zero at \( t = 2 \), and is fairly complicated due to the fifth power term.3. The \( a_x-t \) graph is linear, starting from 9.60 m/s虏 at \( t = 0 \) and dropping to -38.40 m/s虏 at \( t = 2 \).

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Key Concepts

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

position function
In kinematics, the position function, often denoted as \( x(t) \), describes the location of an object relative to a specific point in space as a function of time \( t \). This function tells us where the object is at any given moment.
In the example given, the position function is represented by the equation:
  • \( x(t) = 2.17 + 4.80t^2 - 0.100t^6 \) meters.
This quadratic equation involves coefficients tied to time derivatives, providing insights into how the object's position changes:
  • The constant term (2.17 meters) is the initial position of the car.
  • The \( t^2 \) term shows a quadratic relationship, implying an accelerating motion.
  • The \( t^6 \) term indicates a complex, higher-degree change in position affecting the motion at a much larger scale over time.
Understanding the position function is key to analyzing the motion of the car's bumper over any given time interval, providing a foundation for further exploration of its dynamics.
velocity function
The velocity function, \( v(t) \), represents the rate of change of the position function over time. It describes how fast the position of an object is changing and in what direction.
In this problem, the velocity function is found by taking the derivative of the position function with respect to time:
  • \( v(t) = \frac{dx(t)}{dt} = 9.60t - 0.600t^5 \)
This function indicates two important components:
  • The linear term \( 9.60t \) suggests a basic increase in velocity over time due to constant acceleration.
  • The \( t^5 \) term highlights a complex, nonlinear aspect, which creates fluctuations in velocity.
To find when the car is momentarily at rest, we set \( v(t) = 0 \). Solving for \( t \) helps us determine the critical points where the car's velocity is zero, allowing us to explore its position and acceleration at those moments.
acceleration function
The acceleration function, \( a(t) \), is the derivative of the velocity function and describes how the velocity of an object changes over time. It is a crucial aspect of understanding dynamic movement.
For the car's bumper problem, the acceleration function is given by:
  • \( a(t) = \frac{dv(t)}{dt} = 9.60 - 3.00t^4 \)
Analyzing this expression, we see:
  • The constant term (9.60) implies an initial acceleration level the car experiences.
  • The \( t^4 \) component signifies a substantial shift in acceleration, leading to a decrease over time.
By evaluating acceleration at times where velocity is zero, we can grasp how the acceleration impacts the motion path, providing deeper insights into the car's dynamics at those critical instances.
derivatives in physics
Derivatives are powerful mathematical tools used extensively in physics to describe how quantities change over time. In kinematics, derivatives provide insights into the motion of objects.Here's how derivatives apply to the fundamental concepts involved:
  • The first derivative of the position function \( x(t) \) with respect to time gives us the velocity function \( v(t) \), indicating how the position changes over time.
  • The second derivative, or the derivative of the velocity function, produces the acceleration function \( a(t) \), revealing how velocity shifts over time.
Using derivatives allows physicists and engineers to predict future positions, speeds, and the forces required to achieve certain movements. They are essential for understanding the behavior of dynamic systems and are applied extensively in analyzing real-world problems, like test car dynamics, to optimize performance and ensure safety.

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Most popular questions from this chapter

Look Out Below. Sam heaves a 16-lb shot straight upward, giving it a constant upward acceleration from rest of 35.0 \(\mathrm{m} / \mathrm{s}^{2}\) for 64.0 \(\mathrm{cm} .\) He releases it 2.20 \(\mathrm{m}\) above the ground. You may ignore air resistance. (a) What is the speed of the shot when Sam releases it? (b) How high above the ground does it go? (c) How much time does he have to get out of its way before it returns to the height of the top of his head, 1.83 m above the ground?

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help." When she has fallen for 3.0 s, she hears the echo of her shout from the valley floor below. The speed of sound is 340 \(\mathrm{m} / \mathrm{s}\). (a) How tall is the cliff? (b) If air resistance is neglected, how fast will she be moving just before she hits the ground? (Her actual speed will be less than this, due to air resistance.)

A world-class sprinter accelerates to his maximum speed in 4.0 s. He then maintains this speed for the remainder of a \(100-\mathrm{m}\) race, finishing with a total time of 9.1 \(\mathrm{s}\) . (a) What is the runner's average acceleration during the first 4.0 \(\mathrm{s} ?\) (b) What is his average acceleration during the last 5.1 \(\mathrm{s} ?\) (c) What is his average acceleration for the entire race? (d) Explain why your answer to part (c) is not the average of the answers to parts (a) and (b).

In the vertical jump, an athlete starts from a crouch and jumps upward to reach as high as possible. Even the best athletes spend little more than 1.00 s in the air (their "hang time"). Treat the athlete as a particle and let \(y_{\text { max }}\) be his maximum height above the floor. To explain why he seems to hang in the air, calculate the ratio of the time he is above \(y_{\max } / 2\) to the time it takes him to go from the floor to that height. You may ignore air resistance.

A car travels in the \(+x\) -direction on a straight and level road. For the first 4.00 s of its motion, the average velocity of the car is \(v_{\mathrm{av}-x}=6.25 \mathrm{m} / \mathrm{s} .\) How far does the car travel in 4.00 \(\mathrm{s}\) ?
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