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Chapter 2: Problem 20

The position of the front bumper of a test car under micro-processor 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 (at t=0) and 14.97 m (at t=2). Acceleration: 9.60 m/s虏 (t=0) and -38.40 m/s虏 (t=2). Graph x(t), v(t), a(t) between t=0 and t=2.

Step by step solution

01

Understanding the Problem

We are given a position function of a car bumper, \(x(t)=2.17 \, \mathrm{m}+(4.80 \, \mathrm{m/s}^2) t^2 -(0.100 \, \mathrm{m/s}^6) t^6\). We need to find the position and acceleration when velocity is zero, and graph position, velocity, and acceleration between \(t=0\) and \(t=2.00 \, \mathrm{s}\).
02

Find the Velocity Function

Velocity \(v(t)\) is the derivative of position \(x(t)\). Compute the derivative:\[v(t) = \frac{dx}{dt} = 2(4.80)t - 6(0.100)t^5 = 9.60t - 0.600t^5\]This provides the velocity function \(v(t) = 9.60t - 0.600t^5\).
03

Find the Instants When Velocity is Zero

Set the velocity function to zero and solve for \(t\):\[9.60t - 0.600t^5 = 0 \t(9.60 - 0.600t^4) = 0\]This gives solutions \(t = 0\) and \(t^4 = \frac{9.60}{0.600} = 16\).Thus, \(t^4 = 16\) implies \(t = 卤2\). The relevant \(t\) values are 0 and 2.
04

Calculate Position at Zero Velocity

Using the position function \(x(t)\), plug in \(t = 0\) and \(t = 2\):- For \(t = 0\): \[x(0) = 2.17 + (4.80 \cdot 0^2) - (0.100 \cdot 0^6) = 2.17 \, \mathrm{m}\]- For \(t = 2\): \[x(2) = 2.17 + (4.80 \cdot 2^2) - (0.100 \cdot 2^6) = 2.17 + 19.2 - 6.4 = 14.97 \, \mathrm{m}\]
05

Find the Acceleration Function

Acceleration \(a(t)\) is the derivative of velocity \(v(t)\). Compute the derivative:\[a(t) = \frac{dv}{dt} = 9.60 - 5(0.600)t^4 = 9.60 - 3t^4\]This provides the acceleration function \(a(t) = 9.60 - 3t^4\).
06

Calculate Acceleration at Zero Velocity

Using the acceleration function \(a(t)\), plug in \(t = 0\) and \(t = 2\):- For \(t=0\): \[a(0) = 9.60 - 3(0)^4 = 9.60 \, \mathrm{m/s}^2\]- For \(t=2\): \[a(2) = 9.60 - 3(2)^4 = 9.60 - 48 = -38.40 \, \mathrm{m/s}^2\]
07

Graph Position, Velocity, and Acceleration

Draw the graphs over the interval \([0, 2]\):- **Position** \(x(t)\) is a curve starting at 2.17, increasing to a maximum, then decreasing.- **Velocity** \(v(t)\) is a curve starting at 0, increasing to its maximum, then returning to 0 at \(t=2\).- **Acceleration** \(a(t)\) starts at 9.60 and decreases to a negative value at \(t=2\).

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Key Concepts

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

Derivative in Physics
In physics, derivatives play a crucial role in understanding motion. They help describe how physical quantities change over time. To fathom the movement of objects, like our test car's bumper, we can find derivatives of functions that model its position:
  • **Position to Velocity**: The velocity is the first derivative of the position function. It tells us how quickly the position changes over time.
  • **Velocity to Acceleration**: The acceleration is the second derivative of the position function, or the first derivative of the velocity function. It indicates how the velocity changes over time.
  • The process of differentiation provided us with the velocity \(v(t) = 9.60t - 0.600t^5\) and acceleration \(a(t) = 9.60 - 3t^4\) functions.
This method highlights the derivative's importance in transforming a position-time graph into velocity-time and acceleration-time graphs. Each derivative reveals new insights into the car's motion.
Position Function
The position function is a mathematical expression that describes an object's location relative to a starting point over time. In our problem, the position function is given as:\[x(t) = 2.17 + (4.80)t^2 - (0.100)t^6\]Here's what each part signifies:
  • The constant \(2.17\) meters represents the initial position of the bumper at \(t = 0\).
  • The term \(4.80t^2\) is quadratic, suggesting that the car's position changes with the square of time, often implying accelerated motion.
  • The term \(-0.100t^6\) indicates a change in motion's nature at higher powers of time, suggesting factors like friction or other resistive forces in advanced scenarios.
By evaluating the position function at specific times, especially at when the velocity is zero, we deduce key points in the car's journey.
Acceleration Function
To truly grasp how a car picks up speed or slows down, we study its acceleration function. Acceleration describes the rate of change of velocity. In the given exercise, we derive the acceleration function from the velocity function:\[a(t) = 9.60 - 3t^4\]Understanding this function involves:
  • The constant \(9.60\) \mathrm{m/s}^2\ indicates initial positive acceleration, showing how the bumper starts with a positive rate of speed increase.
  • The term \(-3t^4\) represents the decreasing factor in acceleration as time goes on. At higher times, it overpowers the constant, turning acceleration negative by \(t=2\).
Acceleration becoming negative means the car experiences deceleration at the latter stage of the time interval provided. Evaluating this function at times when velocity hits zero helps us understand critical transition phases in its movement.
Velocity Graph
The velocity graph gives a visual representation of how the speed of the object's motion changes with time. It is derived from the velocity function \(v(t) = 9.60t - 0.600t^5\). Here鈥檚 a deeper dive into understanding this graph:
  • Starts from zero: At initial time \(t = 0\), the car's velocity is zero, as depicted by \(v(0) = 0\).
  • Increases then drops: Initially, the velocity graph ascends, showcasing a rise in speed as \(9.60t\) dominates. However, the \(-0.600t^5\) diminishes velocity, bringing it back to zero at \(t = 2\).
  • The peak of the curve indicates the fastest speed achieved before reducing due to slowing factors, showcasing a typical rise and fall pattern.
Plotting such a graph helps visualize how velocity isn't constant; it varies due to forces impacting the car, capturing the essence of dynamic motion in physics.

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Most popular questions from this chapter

A car's velocity as a function of time is given by \(v_{x}(t)=\) \(\alpha+\beta t^{2},\) where \(\alpha=3.00 \mathrm{m} / \mathrm{s}\) and \(\beta=0.100 \mathrm{m} / \mathrm{s}^{3}\) , (a) Calculate the average acceleration for the time interval \(t=0\) to \(t=5.00 \mathrm{s}\) (b) Calculate the instantaneous acceleration for \(t=0\) and \(t=\) 5.00 \(\mathrm{s}\) . (c) Draw accurate \(v_{x}-t\) and \(a_{x}-t\) graphs for the car's motion between \(t=0\) and \(t=5.00 \mathrm{s} .\)

Earthquake Analysis. Earthquakes produce several types of shock waves. The most well known are the P-waves (P for primary or pressure) and the S-waves (S for secondary or shear). In the earth's crust, the P-waves travel at around 6.5 \(\mathrm{km} / \mathrm{s}\) , while the S-waves move at about 3.5 \(\mathrm{km} / \mathrm{s}\) . The actual speeds vary depending on the type of material they are going through. The time delay between the arrival of these two waves at a seismic recording station tells geologists how far away the earthquake occurred. If the time delay is 33 \(\mathrm{s}\) , how far from the seismic station did the earthquake occur?

Entering the Freeway. A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 20 \(\mathrm{m} / \mathrm{s}(45 \mathrm{mi} / \mathrm{h})\) when it reaches the end of the 120 \(\mathrm{m}\) -long ramp. \((\mathrm{a})\) What is the acceleration of the car? (b) How much time does it take the car to travel the length of the ramp? (c) The traffic on the freeway is moving at a constant speed of 20 \(\mathrm{m} / \mathrm{s}\) . What distance does the traffic travel while the car is moving the length of the ramp?

Freeway Traffic. According to a Scientific American article (May 1990 ), current freeways can sustain about 2400 vehicles per lane per hour in smooth traffic flow at 96 \(\mathrm{km} / \mathrm{h}\) (60 \(\mathrm{mi} / \mathrm{h}\) ). With more vehicles the traffic flow becomes "turbulent" (stop-and- go). (a) If a vehicle is 4.6 \(\mathrm{m}(15 \mathrm{ft})\) long on the average, what is the average spacing between vehicles at the above traffic density? (b) Collision-avoidance automated control systems, which operate by bouncing radar or sonar signals off surrounding vehicles and then accelerate or brake the car when necessary, could greatly reduce the required spacing between vehicles. If the average spacing is 9.2 \(\mathrm{m}\) (two car lengths), how many vehicles per hour can a lane of traffic carry at 96 \(\mathrm{km} / \mathrm{h}\) ?

In an experiment, a shearwater (a seabird) was taken from its nest, flown 5150 \(\mathrm{km}\) away, and released. The bird found its way back to its nest 13.5 days after release. If we place the origin in the nest and extend the \(+x\) -axis to the release point, what was the bird's average velocity in \(\mathrm{m} \mathrm{m} / \mathrm{s}(\mathrm{a})\) for the return flight, and (b) for the whole episode, from leaving the nest to returning?
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