Question
a) Lightning McQueen is originally traveling at 15.0 m/s on a straight, horizontal road. He applies the brakes, causing himself to decelerate uniformly at 4.00 m/s2 until it comes to rest. Calculate the race car's stopping distance. b)IMcQueen was drugged 13 m back to the jail house. What was his displacement? Draw the vectors, Resultant vector in red
Answers
GPT-4o mini
To solve the problem, let's break it down into parts:
### Part (a): Calculate the stopping distance of Lightning McQueen
Given:
- Initial velocity (\( v_i \)) = 15.0 m/s
- Final velocity (\( v_f \)) = 0 m/s (when he comes to rest)
- Deceleration (\( a \)) = -4.00 m/s² (negative since it's deceleration)
We can use the following kinematic equation to find the stopping distance (\( d \)):
\[
v_f^2 = v_i^2 + 2a d
\]
Rearranging to solve for \( d \):
\[
d = \frac{v_f^2 - v_i^2}{2a}
\]
Plugging in the values:
\[
d = \frac{0^2 - (15.0 \, \text{m/s})^2}{2 \times (-4.00 \, \text{m/s}^2)}
\]
\[
d = \frac{-225 \, \text{m}^2/\text{s}^2}{-8.00 \, \text{m/s}^2}
\]
\[
d = \frac{225}{8}
\]
\[
d = 28.125 \, \text{m}
\]
Thus, the stopping distance is **28.1 m** (rounded to three significant figures).
### Part (b): Displacement after being dragged back to the jail house
Lightning McQueen is dragged back \( 13 \, \text{m} \). This displacement will be in the negative direction if we assume that moving back to the starting point (the jail house) is negative.
Assuming the original position is \( 0 \, \text{m} \):
- His initial position (\( x_i \)) = 0 m
- Displacement (\( d \)) = -13 m (dragged back)
Now, the resultant displacement from the original position is:
\[
x_f = x_i + d
\]
\[
x_f = 0 \, \text{m} - 13 \, \text{m} = -13 \, \text{m}
\]
### Vector Diagram
For the vector representation:
- Draw a horizontal line to the right for McQueen's original motion (positive direction).
- From that point, draw a vector 28.1 m to the left to represent the stopping distance.
- Then, from the end of that vector, draw another vector 13 m to the left to represent being dragged back.
The resultant vector (\( R \)) will be the sum of the displacement while stopping (28.1 m left) and the further drag back (13 m left):
\[
R = 28.1 \, \text{m} + 13.0 \, \text{m} = 41.1 \, \text{m}
\]
The final resultant displacement from the starting point is \( -41.1 m \) (in the negative direction), which can be drawn in red on the diagram.
### Summary:
1. Stopping distance: **28.1 m**
2. Displacement after being dragged: **-41.1 m** (to the left).
### Part (a): Calculate the stopping distance of Lightning McQueen
Given:
- Initial velocity (\( v_i \)) = 15.0 m/s
- Final velocity (\( v_f \)) = 0 m/s (when he comes to rest)
- Deceleration (\( a \)) = -4.00 m/s² (negative since it's deceleration)
We can use the following kinematic equation to find the stopping distance (\( d \)):
\[
v_f^2 = v_i^2 + 2a d
\]
Rearranging to solve for \( d \):
\[
d = \frac{v_f^2 - v_i^2}{2a}
\]
Plugging in the values:
\[
d = \frac{0^2 - (15.0 \, \text{m/s})^2}{2 \times (-4.00 \, \text{m/s}^2)}
\]
\[
d = \frac{-225 \, \text{m}^2/\text{s}^2}{-8.00 \, \text{m/s}^2}
\]
\[
d = \frac{225}{8}
\]
\[
d = 28.125 \, \text{m}
\]
Thus, the stopping distance is **28.1 m** (rounded to three significant figures).
### Part (b): Displacement after being dragged back to the jail house
Lightning McQueen is dragged back \( 13 \, \text{m} \). This displacement will be in the negative direction if we assume that moving back to the starting point (the jail house) is negative.
Assuming the original position is \( 0 \, \text{m} \):
- His initial position (\( x_i \)) = 0 m
- Displacement (\( d \)) = -13 m (dragged back)
Now, the resultant displacement from the original position is:
\[
x_f = x_i + d
\]
\[
x_f = 0 \, \text{m} - 13 \, \text{m} = -13 \, \text{m}
\]
### Vector Diagram
For the vector representation:
- Draw a horizontal line to the right for McQueen's original motion (positive direction).
- From that point, draw a vector 28.1 m to the left to represent the stopping distance.
- Then, from the end of that vector, draw another vector 13 m to the left to represent being dragged back.
The resultant vector (\( R \)) will be the sum of the displacement while stopping (28.1 m left) and the further drag back (13 m left):
\[
R = 28.1 \, \text{m} + 13.0 \, \text{m} = 41.1 \, \text{m}
\]
The final resultant displacement from the starting point is \( -41.1 m \) (in the negative direction), which can be drawn in red on the diagram.
### Summary:
1. Stopping distance: **28.1 m**
2. Displacement after being dragged: **-41.1 m** (to the left).