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Chapter 6: Problem 20

A car going \(80 \mathrm{ft} / \mathrm{sec}\) (about \(55 \mathrm{mph}\) ) brakes to a stop in five seconds. Assume the deceleration is constant. (a) Graph the velocity against time, \(t,\) for \(0 \leq t \leq 5\) seconds. (b) Represent, as an area on the graph, the total distance traveled from the time the brakes are applied until the car comes to a stop. (c) Find this area and hence the distance traveled. (d) Now find the total distance traveled using anti-differentiation.

Short Answer

Expert verified
The car travels 200 feet during the braking period.

Step by step solution

01

Understand the problem

We need to find a way to model the velocity of a car decreasing linearly from 80 feet per second to 0 over 5 seconds, calculate the distance it travels during this time, and sketch the graph of velocity against time.
02

Write the velocity function

Since the car decelerates at a constant rate, its velocity as a function of time is linear. We can write it as:\[ v(t) = 80 - 16t \]where the initial velocity is 80 ft/s and after 5 seconds (when the car stops) the velocity is 0 ft/s: \[ 80 - 16 \times 5 = 0 \].
03

Graph the velocity function

On a graph with time (t) on the x-axis and velocity (v) on the y-axis, plot the line defined by the equation \( v(t) = 80 - 16t \) from \( t = 0 \) to \( t = 5 \). The line will start at (0, 80) and end at (5, 0), forming a straight line.
04

Identify the area under the graph

The area under the velocity-time graph represents the distance traveled. Since it's a straight line decreasing to zero, the area forms a right triangle with a base of 5 (seconds) and a height of 80 (ft/s).
05

Calculate the distance using the graph

The area of the triangle is:\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 80 = 200 \text{ feet} \]. The total distance traveled is 200 feet.
06

Calculate the distance using anti-differentiation

To find the distance traveled using anti-differentiation, integrate the velocity function: \[ s(t) = \int (80 - 16t) \, dt \] Calculating this integral:\[ s(t) = 80t - 8t^2 + C \]Since the initial position is 0 when t=0, C=0.Evaluate from t=0 to t=5:\[ s(5) - s(0) = (80 \times 5 - 8 \times 5^2) - (0) = 400 - 200 = 200 \text{ feet} \].

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Key Concepts

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

Understanding the Velocity-Time Graph
A velocity-time graph provides a visual representation of how velocity changes over time. In this particular problem, the car starts with a velocity of 80 feet per second and gradually decreases to 0 feet per second over a span of 5 seconds. This happens because of constant deceleration.
The graph of this scenario is a straight line that slopes downward from the starting point (0, 80) to the end point (5, 0).
  • The time (in seconds) is plotted on the x-axis.
  • The velocity (in feet per second) is on the y-axis.
This aids students visually by making it easier to interpret the change in velocity over time.
Calculating Distance from the Velocity-Time Graph
The total distance traveled during the deceleration is represented by the area under the velocity-time graph. For our graph, this area forms a right triangle with a base going from 0 to 5 seconds and a height up to 80 feet per second.

To find the area of a right triangle, use the formula:
  • Area = \( \frac{1}{2} \times \text{base} \times \text{height} \)
Applying this formula gives:
  • Area = \( \frac{1}{2} \times 5 \times 80 \)
This equals 200 feet, which is the total distance traveled by the car during deceleration.
Using Anti-Differentiation to Find Distance
Anti-differentiation, or integration, is a powerful tool in calculus used to calculate areas under curves, such as velocity-time graphs. Here, we need to find the distance by integrating the velocity function:

Our velocity function is \( v(t) = 80 - 16t \).
  • To integrate, we compute \( s(t) = \int (80 - 16t) \, dt \).
  • This results in \( s(t) = 80t - 8t^2 + C \).
Given that the car starts at position 0 when \( t = 0 \), we set \( C = 0 \).
Evaluating this from \( t = 0 \) to \( t = 5 \), the computation shows:
  • \( s(5) = 400 - 200 = 200 \; \text{feet} \).
This confirms the distance traveled as 200 feet.
Concept of Constant Deceleration
Constant deceleration occurs when the velocity of an object reduces at a steady rate over time. In our problem, this means the car's speed decreases linearly from 80 feet per second to zero within 5 seconds.

This behavior is modeled by the linear function: \( v(t) = 80 - 16t \). Here:
  • 80 is the initial velocity.
  • -16 is the constant rate of deceleration.
With constant deceleration, the velocity-time relationship is a straight line on the graph, simplifying the calculation of distances and demonstrating a straightforward decrease in speed.

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