Two motorcycles are travelling due east with different velocities. However, four seconds later, they have the same velocity, During this four-second iterval, motorcycle A has an average acceleration of `2.0 m//s^(2)` due east, while motorcycle B has an average acceleration of `4.0 m//s^(2)` due east. By how much did the speeds differ at the beginning of the four-second interval, and which motorcycle was moving faster ?

Options

A. `-2.0` m/s
B. `-6.0` m/s
C. `+8.0` m/s
D. `+1.0` m/s

Answer

To solve the problem, we need to determine the initial speeds of both motorcycles (A and B) before the four-second interval and find out how much they differ. We also need to identify which motorcycle was moving faster initially.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Motorcycle A has an average acceleration of \(2.0 \, \text{m/s}^2\) due east.
- Motorcycle B has an average acceleration of \(4.0 \, \text{m/s}^2\) due east.
- After 4 seconds, both motorcycles have the same final velocity.

2. **Using the Formula for Acceleration**:
- The formula for average acceleration is given by:
\[ a = \frac{\Delta v}{\Delta t} \]
- Here, \(\Delta v\) is the change in velocity, and \(\Delta t\) is the time interval.

3. **Setting Up the Equations**:
- For Motorcycle A:
\[ a_A = \frac{v - v_1}{4} \]
where \(v_1\) is the initial velocity of Motorcycle A, and \(v\) is the final velocity after 4 seconds.
- Given \(a_A = 2.0 \, \text{m/s}^2\):
\[ 2 = \frac{v - v_1}{4} \]
Rearranging gives:
\[ v - v_1 = 8 \quad \text{(Equation 1)} \]

- For Motorcycle B:
\[ a_B = \frac{v - v_2}{4} \]
where \(v_2\) is the initial velocity of Motorcycle B.
- Given \(a_B = 4.0 \, \text{m/s}^2\):
\[ 4 = \frac{v - v_2}{4} \]
Rearranging gives:
\[ v - v_2 = 16 \quad \text{(Equation 2)} \]

4. **Finding the Difference in Initial Speeds**:
- Now, we have two equations:
- From Equation 1: \(v = v_1 + 8\)
- From Equation 2: \(v = v_2 + 16\)
- Setting the two expressions for \(v\) equal:
\[ v_1 + 8 = v_2 + 16 \]
- Rearranging gives:
\[ v_1 - v_2 = 16 - 8 \]
\[ v_1 - v_2 = 8 \]
- This means that the initial speed of Motorcycle A was \(8 \, \text{m/s}\) greater than that of Motorcycle B.

5. **Determining Which Motorcycle Was Faster**:
- Since \(v_1 - v_2 = 8\), it follows that:
\[ v_1 > v_2 \]
- Therefore, Motorcycle A was moving faster than Motorcycle B at the beginning of the four-second interval.

### Final Answers:
- The speeds differed by \(8 \, \text{m/s}\).
- Motorcycle A was moving faster than Motorcycle B.