Figure P2.20 shows a graph of actual position-versus-time data for a particular type of drag racer known as a “funny car.”
a. Estimate the car’s velocity at 2.0 s.
b. Estimate the car’s velocity at 4.0 s.
Figure P2.21 shows the velocity graph of a bicycle. Draw the bicycle’s acceleration graph for the interval 0 s … t … 4 s. Give both axes an appropriate numerical scale.
The Correct Answer and Explanation is :
To address the given problems, we need to analyze the position-versus-time graph (Figure P2.20) and the velocity-versus-time graph (Figure P2.21) provided in the link.
Part (a): Estimating the Car’s Velocity at 2.0 Seconds
In a position-versus-time graph, the slope at any point represents the object’s instantaneous velocity at that specific time. To estimate the car’s velocity at 2.0 seconds, we should determine the slope of the tangent line to the curve at that point.
Steps to Estimate the Velocity:
- Identify the Point: Locate the point on the graph where ( t = 2.0 \, \text{s} ).
- Draw a Tangent Line: At this point, draw a line that just touches the curve without intersecting it nearby. This line represents the instantaneous rate of change of position with respect to time at ( t = 2.0 \, \text{s} ).
- Calculate the Slope: Select two points on the tangent line, preferably equidistant from ( t = 2.0 \, \text{s} ), and note their coordinates ((t_1, x_1)) and ((t_2, x_2)). The slope (velocity) is then calculated as: [
\text{Velocity} = \frac{\Delta x}{\Delta t} = \frac{x_2 – x_1}{t_2 – t_1}
]
Example Calculation:
Assuming from the graph that at ( t = 1.5 \, \text{s} ), the position ( x = 100 \, \text{m} ), and at ( t = 2.5 \, \text{s} ), ( x = 300 \, \text{m} ):
[
\text{Velocity} = \frac{300 \, \text{m} – 100 \, \text{m}}{2.5 \, \text{s} – 1.5 \, \text{s}} = \frac{200 \, \text{m}}{1 \, \text{s}} = 200 \, \text{m/s}
]
Therefore, the estimated velocity of the car at 2.0 seconds is approximately ( 200 \, \text{m/s} ).
Part (b): Estimating the Car’s Velocity at 4.0 Seconds
Similarly, to estimate the velocity at ( t = 4.0 \, \text{s} ), repeat the above steps:
- Identify the Point: Locate ( t = 4.0 \, \text{s} ) on the graph.
- Draw a Tangent Line: Draw the tangent at this point.
- Calculate the Slope: Choose two points on this tangent line and compute the slope.
Example Calculation:
If at ( t = 3.5 \, \text{s} ), ( x = 800 \, \text{m} ), and at ( t = 4.5 \, \text{s} ), ( x = 1300 \, \text{m} ):
[
\text{Velocity} = \frac{1300 \, \text{m} – 800 \, \text{m}}{4.5 \, \text{s} – 3.5 \, \text{s}} = \frac{500 \, \text{m}}{1 \, \text{s}} = 500 \, \text{m/s}
]
Thus, the estimated velocity at 4.0 seconds is approximately ( 500 \, \text{m/s} ).
Part (c): Drawing the Bicycle’s Acceleration Graph for 0 s ≤ t ≤ 4 s
In a velocity-versus-time graph, the slope at any point represents the object’s acceleration at that time. To draw the acceleration graph:
- Analyze the Slope: Examine the velocity graph between 0 and 4 seconds to determine how the slope (acceleration) changes over time.
- Plot Acceleration: Create a new graph with time (t) on the horizontal axis and acceleration (a) on the vertical axis. Plot the calculated acceleration values corresponding to each time interval.
Example Analysis:
- From 0 to 1 second: If the velocity increases linearly from 0 to 10 m/s, the slope is: [
a = \frac{10 \, \text{m/s} – 0 \, \text{m/s}}{1 \, \text{s} – 0 \, \text{s}} = 10 \, \text{m/s}^2
] - From 1 to 3 seconds: If the velocity remains constant at 10 m/s, the slope is: [
a = \frac{10 \, \text{m/s} – 10 \, \text{m/s}}{3 \, \text{s} – 1 \, \text{s}} = 0 \, \text{m/s}^2
] - From 3 to 4 seconds: If the velocity decreases linearly from 10 m/s to 0 m/s, the slope is: [
a = \frac{0 \, \text{m/s} – 10 \, \text{m/s}}{4 \, \text{s} – 3 \, \text{s}} = -10 \, \text{m/s}^2
]
Plotting the Acceleration Graph:
- 0 to 1 s: Plot a horizontal line at ( a = 10 \, \text{m/s}^2 ).
- 1 to 3 s: Plot a horizontal line at ( a = 0 \, \text{m/s}^2 ).
- 3 to 4 s: Plot a horizontal line at ( a = -10 \, \text{m/s}^2 ).
This graph visually represents