**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?
Displacement, Velocity and Acceleration
In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.
Linear Displacement
The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.
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)
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