91Ó°ÊÓ

Understanding how to convert units is crucial when dealing with problems in physics. In this exercise, we need to convert miles and yards to meters, which are the standard units used in physics calculations.
To convert miles to meters:

• We know that 1 mile equals approximately 1609.34 meters.

For converting yards to meters:

• 500 yards are equivalent to 457.2 meters since 1 yard equals roughly 0.9144 meters.

These conversions ensure that all distances are in the same units, simplifying the calculations. Consistency in units is vital for avoiding errors and ensuring that the formulas used yield correct and coherent results. This kind of conversion becomes particularly important for solving kinematic equations.

The concept of acceleration is about how quickly an object's speed increases. In this problem, we explore how fast the student can possibly accelerate.
First, we identify the student's maximum acceleration as 0.15 m/s². We then need to determine if she can accelerate from rest (starting speed) to the minimum speed required to finish her run in time.
Using the formula:\[ t = \frac{v}{a_{max}} \]where

• \(t\) is the time needed,
• \(v\) is the final speed (3.81 m/s, calculated from the given distance and remaining time),
• and \(a_{max}\) is the maximum acceleration.

We found out that she would need 25.4 seconds to achieve this speed, which is more than the available 120 seconds. Hence, the maximum acceleration is not sufficient in this scenario.
Acceleration calculations are essential for understanding the feasibility of speeding up to meet time constraints given certain distance and speed conditions.

The relationship between time, speed, and distance is a fundamental concept in kinematics. This relation helps us determine how long it takes to travel a certain distance at a given speed or conversely, the speed needed to cover a distance in a certain timeframe.
The equation used here is:\[ v = \frac{d}{T} \]where

• \(v\) is speed,
• \(d\) is distance remaining (457.2 meters for this case),
• and \(T\) is the remaining time (120 seconds in this problem).

This formula tells us that to cover the remaining distance in the time she has left, the student needs a constant speed of 3.81 m/s.
This calculation is crucial because it sets the target speed, which is then used to evaluate if her maximum acceleration can help her reach this speed. In kinematics, understanding the interaction between these three elements—time, speed, and distance—allows for solving a broad array of motion-related problems.