Answer :
Certainly! Let's break down the solution step-by-step:
### Part (a): Plot the graph of the car's position versus time
The position function given is:
[tex]\[ x(t) = (5.0 \, \text{m/s}) t + (0.75 \, \text{m/s}^3) t^3 \][/tex]
1. Creating a Table of Values:
We need to calculate [tex]\( x \)[/tex] for various values of [tex]\( t \)[/tex] from 0 to 4 seconds to plot the graph. Here are some sample calculations:
- When [tex]\( t = 0 \)[/tex]:
[tex]\[ x(0) = 5.0 \times 0 + 0.75 \times 0^3 = 0 \][/tex]
- When [tex]\( t = 1 \)[/tex]:
[tex]\[ x(1) = 5.0 \times 1 + 0.75 \times 1^3 = 5.0 + 0.75 = 5.75 \, \text{m} \][/tex]
- When [tex]\( t = 2 \)[/tex]:
[tex]\[ x(2) = 5.0 \times 2 + 0.75 \times 2^3 = 10.0 + 6.0 = 16.0 \, \text{m} \][/tex]
- When [tex]\( t = 3 \)[/tex]:
[tex]\[ x(3) = 5.0 \times 3 + 0.75 \times 3^3 = 15.0 + 20.25 = 35.25 \, \text{m} \][/tex]
- When [tex]\( t = 4 \)[/tex]:
[tex]\[ x(4) = 5.0 \times 4 + 0.75 \times 4^3 = 20.0 + 48.0 = 68.0 \, \text{m} \][/tex]
2. Plotting the Graph:
- (Time [tex]\( t \)[/tex] on the x-axis)
- (Position [tex]\( x(t) \)[/tex] on the y-axis)
- Plot points for [tex]\( t \)[/tex] from 0 to 4 seconds.
- Connect the points smoothly because the relationship [tex]\( x(t) \)[/tex] is continuous and smooth.
### Part (b): Determine the instantaneous velocity at [tex]\( t = 4.0 \)[/tex] seconds
To approximate the instantaneous velocity, we use the average velocities over smaller and smaller time intervals ([tex]\( \Delta t \)[/tex]) around [tex]\( t = 4.0 \)[/tex].
For instantaneous velocity at [tex]\( t = 4.0 \)[/tex]:
[tex]\[ v(t) = \frac{x(t+\Delta t) - x(t)}{\Delta t} \][/tex]
#### For [tex]\( \Delta t = 0.40 \)[/tex]:
[tex]\[ v(4) = \frac{x(4.4) - x(4)}{0.40} \][/tex]
First, calculate [tex]\( x(4.4) \)[/tex]:
[tex]\[ x(4.4) = 5.0 \times 4.4 + 0.75 \times (4.4)^3 = 22 + 0.75 \times 85.184 = 22 + 63.888 = 85.888 \, \text{m} \][/tex]
Therefore,
[tex]\[ v(4) = \frac{85.888 - 68.0}{0.40} = \frac{17.888}{0.40} = 44.72 \, \text{m/s} \][/tex]
#### For [tex]\( \Delta t = 0.20 \)[/tex]:
[tex]\[ v(4) = \frac{x(4.2) - x(4)}{0.20} \][/tex]
First, calculate [tex]\( x(4.2) \)[/tex]:
[tex]\[ x(4.2) = 5.0 \times 4.2 + 0.75 \times (4.2)^3 = 21 + 0.75 \times 74.088 = 21 + 55.566 = 76.566 \, \text{m} \][/tex]
Therefore,
[tex]\[ v(4) = \frac{76.566 - 68.0}{0.20} = \frac{8.566}{0.20} = 42.83 \, \text{m/s} \][/tex]
#### For [tex]\( \Delta t = 0.10 \)[/tex]:
[tex]\[ v(4) = \frac{x(4.1) - x(4)}{0.10} \][/tex]
First, calculate [tex]\( x(4.1) \)[/tex]:
[tex]\[ x(4.1) = 5.0 \times 4.1 + 0.75 \times (4.1)^3 = 20.5 + 0.75 \times 68.921 = 20.5 + 51.69075 = 72.19075 \, \text{m} \][/tex]
Therefore,
[tex]\[ v(4) = \frac{72.19075 - 68.0}{0.10} = \frac{4.19075}{0.10} = 41.91 \, \text{m/s} \][/tex]
### Part (c): Average velocity during the first 4.0 seconds
The formula for average velocity ([tex]\( \bar{v} \)[/tex]) over a time interval is:
[tex]\[ \bar{v} = \frac{x(t_{\text{final}}) - x(t_{\text{initial}})}{t_{\text{final}} - t_{\text{initial}}} \][/tex]
For [tex]\( t = 0 \)[/tex] to [tex]\( t = 4 \)[/tex]:
[tex]\[ \bar{v} = \frac{x(4) - x(0)}{4 - 0} = \frac{68.0 - 0}{4} = 17.0 \, \text{m/s} \][/tex]
### Comparison
- The average velocity over the first 4.0 seconds (17.0 m/s) is significantly lower than the instantaneous velocities at [tex]\( t = 4.0 \)[/tex].
- The smaller the time interval ([tex]\(\Delta t\)[/tex]), the more accurate the approximation of instantaneous velocity. The values approximately calculated for [tex]\( v(4) \)[/tex] were 44.72 m/s, 42.83 m/s, and 41.91 m/s with increasingly smaller intervals.
- As expected, given the cubic term in the position function, the velocity increases over time. Thus, the instantaneous velocity at the end of the interval is much higher than the average over the entire interval, which is why the average is lower.
### Part (a): Plot the graph of the car's position versus time
The position function given is:
[tex]\[ x(t) = (5.0 \, \text{m/s}) t + (0.75 \, \text{m/s}^3) t^3 \][/tex]
1. Creating a Table of Values:
We need to calculate [tex]\( x \)[/tex] for various values of [tex]\( t \)[/tex] from 0 to 4 seconds to plot the graph. Here are some sample calculations:
- When [tex]\( t = 0 \)[/tex]:
[tex]\[ x(0) = 5.0 \times 0 + 0.75 \times 0^3 = 0 \][/tex]
- When [tex]\( t = 1 \)[/tex]:
[tex]\[ x(1) = 5.0 \times 1 + 0.75 \times 1^3 = 5.0 + 0.75 = 5.75 \, \text{m} \][/tex]
- When [tex]\( t = 2 \)[/tex]:
[tex]\[ x(2) = 5.0 \times 2 + 0.75 \times 2^3 = 10.0 + 6.0 = 16.0 \, \text{m} \][/tex]
- When [tex]\( t = 3 \)[/tex]:
[tex]\[ x(3) = 5.0 \times 3 + 0.75 \times 3^3 = 15.0 + 20.25 = 35.25 \, \text{m} \][/tex]
- When [tex]\( t = 4 \)[/tex]:
[tex]\[ x(4) = 5.0 \times 4 + 0.75 \times 4^3 = 20.0 + 48.0 = 68.0 \, \text{m} \][/tex]
2. Plotting the Graph:
- (Time [tex]\( t \)[/tex] on the x-axis)
- (Position [tex]\( x(t) \)[/tex] on the y-axis)
- Plot points for [tex]\( t \)[/tex] from 0 to 4 seconds.
- Connect the points smoothly because the relationship [tex]\( x(t) \)[/tex] is continuous and smooth.
### Part (b): Determine the instantaneous velocity at [tex]\( t = 4.0 \)[/tex] seconds
To approximate the instantaneous velocity, we use the average velocities over smaller and smaller time intervals ([tex]\( \Delta t \)[/tex]) around [tex]\( t = 4.0 \)[/tex].
For instantaneous velocity at [tex]\( t = 4.0 \)[/tex]:
[tex]\[ v(t) = \frac{x(t+\Delta t) - x(t)}{\Delta t} \][/tex]
#### For [tex]\( \Delta t = 0.40 \)[/tex]:
[tex]\[ v(4) = \frac{x(4.4) - x(4)}{0.40} \][/tex]
First, calculate [tex]\( x(4.4) \)[/tex]:
[tex]\[ x(4.4) = 5.0 \times 4.4 + 0.75 \times (4.4)^3 = 22 + 0.75 \times 85.184 = 22 + 63.888 = 85.888 \, \text{m} \][/tex]
Therefore,
[tex]\[ v(4) = \frac{85.888 - 68.0}{0.40} = \frac{17.888}{0.40} = 44.72 \, \text{m/s} \][/tex]
#### For [tex]\( \Delta t = 0.20 \)[/tex]:
[tex]\[ v(4) = \frac{x(4.2) - x(4)}{0.20} \][/tex]
First, calculate [tex]\( x(4.2) \)[/tex]:
[tex]\[ x(4.2) = 5.0 \times 4.2 + 0.75 \times (4.2)^3 = 21 + 0.75 \times 74.088 = 21 + 55.566 = 76.566 \, \text{m} \][/tex]
Therefore,
[tex]\[ v(4) = \frac{76.566 - 68.0}{0.20} = \frac{8.566}{0.20} = 42.83 \, \text{m/s} \][/tex]
#### For [tex]\( \Delta t = 0.10 \)[/tex]:
[tex]\[ v(4) = \frac{x(4.1) - x(4)}{0.10} \][/tex]
First, calculate [tex]\( x(4.1) \)[/tex]:
[tex]\[ x(4.1) = 5.0 \times 4.1 + 0.75 \times (4.1)^3 = 20.5 + 0.75 \times 68.921 = 20.5 + 51.69075 = 72.19075 \, \text{m} \][/tex]
Therefore,
[tex]\[ v(4) = \frac{72.19075 - 68.0}{0.10} = \frac{4.19075}{0.10} = 41.91 \, \text{m/s} \][/tex]
### Part (c): Average velocity during the first 4.0 seconds
The formula for average velocity ([tex]\( \bar{v} \)[/tex]) over a time interval is:
[tex]\[ \bar{v} = \frac{x(t_{\text{final}}) - x(t_{\text{initial}})}{t_{\text{final}} - t_{\text{initial}}} \][/tex]
For [tex]\( t = 0 \)[/tex] to [tex]\( t = 4 \)[/tex]:
[tex]\[ \bar{v} = \frac{x(4) - x(0)}{4 - 0} = \frac{68.0 - 0}{4} = 17.0 \, \text{m/s} \][/tex]
### Comparison
- The average velocity over the first 4.0 seconds (17.0 m/s) is significantly lower than the instantaneous velocities at [tex]\( t = 4.0 \)[/tex].
- The smaller the time interval ([tex]\(\Delta t\)[/tex]), the more accurate the approximation of instantaneous velocity. The values approximately calculated for [tex]\( v(4) \)[/tex] were 44.72 m/s, 42.83 m/s, and 41.91 m/s with increasingly smaller intervals.
- As expected, given the cubic term in the position function, the velocity increases over time. Thus, the instantaneous velocity at the end of the interval is much higher than the average over the entire interval, which is why the average is lower.
