**23. Two motorcycles are traveling due east with different velocities. However, four seconds later, they have the same velocity. During this four-second interval, cycle A has an average acceleration of 2.0 m/s² due east, while cycle B has an average acceleration of 4.0 m/s2 due east. By how much did the speeds differ at the beginning of the four-second inter- val, and which motorcycle was moving faster?
**23. Two motorcycles are traveling due east with different velocities. However, four seconds later, they have the same velocity. During this four-second interval, cycle A has an average acceleration of 2.0 m/s² due east, while cycle B has an average acceleration of 4.0 m/s2 due east. By how much did the speeds differ at the beginning of the four-second inter- val, and which motorcycle was moving faster?
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Physics Question #23
![**Problem 23: Motion of Two Motorcycles**
**Scenario:**
Two motorcycles are traveling due east but have different initial velocities. Four seconds later, they are moving at the same velocity.
- During this four-second interval:
- Motorcycle A has an average acceleration of 2.0 m/s² due east.
- Motorcycle B has an average acceleration of 4.0 m/s² due east.
**Question:**
*By how much did the speeds differ at the beginning of the four-second interval, and which motorcycle was moving faster initially?*
**Analysis:**
1. Let's denote the initial velocities of motorcycle A and motorcycle B as \( v_{A_0} \) and \( v_{B_0} \) respectively.
2. After four seconds, they have reached the same velocity, say \( v_f \).
For Motorcycle A:
\[
v_f = v_{A_0} + (2.0 \, \frac{m}{s^2}) \times 4 \, s
\]
For Motorcycle B:
\[
v_f = v_{B_0} + (4.0 \, \frac{m}{s^2}) \times 4 \, s
\]
Since both reach the same final velocity \( v_f \):
\[
v_{A_0} + 8 = v_{B_0} + 16
\]
Therefore:
\[
v_{B_0} = v_{A_0} - 8
\]
**Conclusion:**
Motorcycle A's initial speed was faster by 8 m/s compared to Motorcycle B's initial speed.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2c8c9d8-cb7d-4bb3-b241-ba372e4ba26e%2Ff6798b5f-43af-426a-8017-e056bca2423c%2Fsmhie7k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 23: Motion of Two Motorcycles**
**Scenario:**
Two motorcycles are traveling due east but have different initial velocities. Four seconds later, they are moving at the same velocity.
- During this four-second interval:
- Motorcycle A has an average acceleration of 2.0 m/s² due east.
- Motorcycle B has an average acceleration of 4.0 m/s² due east.
**Question:**
*By how much did the speeds differ at the beginning of the four-second interval, and which motorcycle was moving faster initially?*
**Analysis:**
1. Let's denote the initial velocities of motorcycle A and motorcycle B as \( v_{A_0} \) and \( v_{B_0} \) respectively.
2. After four seconds, they have reached the same velocity, say \( v_f \).
For Motorcycle A:
\[
v_f = v_{A_0} + (2.0 \, \frac{m}{s^2}) \times 4 \, s
\]
For Motorcycle B:
\[
v_f = v_{B_0} + (4.0 \, \frac{m}{s^2}) \times 4 \, s
\]
Since both reach the same final velocity \( v_f \):
\[
v_{A_0} + 8 = v_{B_0} + 16
\]
Therefore:
\[
v_{B_0} = v_{A_0} - 8
\]
**Conclusion:**
Motorcycle A's initial speed was faster by 8 m/s compared to Motorcycle B's initial speed.
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