Question
What is the solution to a physics problem involving relative motion, where one person is jogging at a constant speed and another person is accelerating to catch up?
Answers
GPT-4o mini
To solve a physics problem involving relative motion where one person is jogging at a constant speed and another person is accelerating to catch up, you'll typically follow these steps:
### Step 1: Define the Variables
- Let \( v_j \) be the constant speed of the jogger (in meters per second).
- Let \( a \) be the constant acceleration of the person trying to catch up (in meters per second squared).
- Let \( d_0 \) be the initial distance between the jogger and the accelerating person (in meters).
- Let \( t \) be the time (in seconds) after the person starts accelerating.
### Step 2: Write the Equations of Motion
1. **Position of the Jogger**:
The jogger moves at a constant speed, so their position as a function of time is:
\[
x_j(t) = v_j t
\]
2. **Position of the Accelerating Person**:
The person who is accelerating can be described by the equation of motion:
\[
x_a(t) = \frac{1}{2} a t^2 + d_0
\]
Note: The initial position of the accelerating person can be considered as \( d_0 \) (the distance behind the jogger).
### Step 3: Set the Positions Equal
To find when the accelerating person catches up with the jogger, set their positions equal:
\[
v_j t = \frac{1}{2} a t^2 + d_0
\]
### Step 4: Rearrange the Equation
Rearranging gives:
\[
\frac{1}{2} a t^2 - v_j t + d_0 = 0
\]
This is a quadratic equation in the form \( At^2 + Bt + C = 0 \):
- \( A = \frac{1}{2} a \)
- \( B = -v_j \)
- \( C = d_0 \)
### Step 5: Solve the Quadratic Equation
You can use the quadratic formula to find \( t \):
\[
t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} = \frac{v_j \pm \sqrt{(-v_j)^2 - 4 \cdot \frac{1}{2} a \cdot d_0}}{2 \cdot \frac{1}{2} a}
\]
Simplifying gives:
\[
t = \frac{v_j \pm \sqrt{v_j^2 - 2 a d_0}}{a}
\]
### Step 6: Analyze the Solution
- Choose the positive value of \( t \) since time cannot be negative.
- Verify the discriminant \( (v_j^2 - 2 a d_0) \) is non-negative to ensure a real solution exists.
### Step 7: Conclusion
This value of \( t \) gives you the time it takes for the accelerating person to catch up with the jogger. Further calculations can be done to find their positions at the time of catching up or to analyze motion thereafter.
### Example
If \( v_j = 5 \, \text{m/s} \), \( a = 2 \, \text{m/s}^2 \), and \( d_0 = 10 \, \text{m} \):
1. Set up the quadratic equation:
\[
\frac{1}{2} (2)t^2 - 5t + 10 = 0 \hspace{10mm} (t^2 - 5t + 10 = 0)
\]
2. Solve this using the quadratic formula to find \( t \).
By following these steps, you can systematically approach and solve relative motion problems involving different types of motion for multiple participants.
### Step 1: Define the Variables
- Let \( v_j \) be the constant speed of the jogger (in meters per second).
- Let \( a \) be the constant acceleration of the person trying to catch up (in meters per second squared).
- Let \( d_0 \) be the initial distance between the jogger and the accelerating person (in meters).
- Let \( t \) be the time (in seconds) after the person starts accelerating.
### Step 2: Write the Equations of Motion
1. **Position of the Jogger**:
The jogger moves at a constant speed, so their position as a function of time is:
\[
x_j(t) = v_j t
\]
2. **Position of the Accelerating Person**:
The person who is accelerating can be described by the equation of motion:
\[
x_a(t) = \frac{1}{2} a t^2 + d_0
\]
Note: The initial position of the accelerating person can be considered as \( d_0 \) (the distance behind the jogger).
### Step 3: Set the Positions Equal
To find when the accelerating person catches up with the jogger, set their positions equal:
\[
v_j t = \frac{1}{2} a t^2 + d_0
\]
### Step 4: Rearrange the Equation
Rearranging gives:
\[
\frac{1}{2} a t^2 - v_j t + d_0 = 0
\]
This is a quadratic equation in the form \( At^2 + Bt + C = 0 \):
- \( A = \frac{1}{2} a \)
- \( B = -v_j \)
- \( C = d_0 \)
### Step 5: Solve the Quadratic Equation
You can use the quadratic formula to find \( t \):
\[
t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} = \frac{v_j \pm \sqrt{(-v_j)^2 - 4 \cdot \frac{1}{2} a \cdot d_0}}{2 \cdot \frac{1}{2} a}
\]
Simplifying gives:
\[
t = \frac{v_j \pm \sqrt{v_j^2 - 2 a d_0}}{a}
\]
### Step 6: Analyze the Solution
- Choose the positive value of \( t \) since time cannot be negative.
- Verify the discriminant \( (v_j^2 - 2 a d_0) \) is non-negative to ensure a real solution exists.
### Step 7: Conclusion
This value of \( t \) gives you the time it takes for the accelerating person to catch up with the jogger. Further calculations can be done to find their positions at the time of catching up or to analyze motion thereafter.
### Example
If \( v_j = 5 \, \text{m/s} \), \( a = 2 \, \text{m/s}^2 \), and \( d_0 = 10 \, \text{m} \):
1. Set up the quadratic equation:
\[
\frac{1}{2} (2)t^2 - 5t + 10 = 0 \hspace{10mm} (t^2 - 5t + 10 = 0)
\]
2. Solve this using the quadratic formula to find \( t \).
By following these steps, you can systematically approach and solve relative motion problems involving different types of motion for multiple participants.