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If you run at \(4.0 \mathrm{~m} / \mathrm{s}\) for \(100 \mathrm{~m},\) then at \(5.0 \mathrm{~m} / \mathrm{s}\) for another \(100 \mathrm{~m},\) what's your average speed?

Short Answer

Expert verified
The average speed is approximately 4.44 m/s.

Step by step solution

01

Calculate the time for each segment

To find the time taken for each segment, use the formula \( t = \frac{d}{s} \), where \( t \) is time, \( d \) is distance, and \( s \) is speed. For the first segment, \( d = 100 \, \text{m} \) and \( s = 4.0 \, \text{m/s} \), so \( t_1 = \frac{100}{4.0} = 25 \, \text{s} \). For the second segment, \( d = 100 \, \text{m} \) and \( s = 5.0 \, \text{m/s} \), so \( t_2 = \frac{100}{5.0} = 20 \, \text{s} \).
02

Calculate the total time and distance

The total distance run is the sum of the distances of the two segments, \( D = 100 \, \text{m} + 100 \, \text{m} = 200 \, \text{m} \). The total time taken is the sum of the times for each segment, \( T = 25 \, \text{s} + 20 \, \text{s} = 45 \, \text{s} \).
03

Calculate the average speed

The average speed \( v_\text{avg} \) is given by the total distance divided by the total time, \( v_\text{avg} = \frac{D}{T} = \frac{200}{45} \approx 4.44 \, \text{m/s} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematics
Kinematics is an essential concept in physics that focuses on the motion of objects without considering the forces causing the motion. It involves analyzing speed, velocity, and acceleration among other parameters. In our exercise, we specifically deal with the concept of speed and how it changes over time and distance.

When solving problems in kinematics, it's vital to understand the fundamental parameters:
  • Speed: The rate at which an object covers distance. It is a scalar quantity, which means it has magnitude but no direction. For example, running at 4 m/s is a speed.
  • Velocity: Similar to speed but with direction. However, since our current focus is average speed, we won't delve deeply into velocity.
  • Time: Refers to the duration taken to move from one point to another.
  • Distance: The total path traveled by the object, a scalar quantity as well.
By carefully dissecting these terms, one can develop a more accurate understanding of a moving object's behavior. Here, kinematics helped us find how long the person took to complete two segments of running, leading us to calculate the average speed.
Problem Solving
Problem-solving in physics requires a structured approach, especially in kinematics.

To address the given problem, we started by identifying which formulas are needed. Generally:
  • Identify Known Variables: We knew the distances and speeds of each segment.
  • Determine Unknowns: Here, it was the average speed.
  • Apply Relevant Formulas: We used the formula for time, \( t = \frac{d}{s} \), to find the time for each run segment.
  • Solve Incrementally: First, calculate the time for each segment, then add to find the total time, and eventually use total distance and time to find the average speed.
By breaking down the problem into smaller, more manageable calculations, one ensures accuracy and gains confidence in finding the solution. This step-by-step technique is crucial in tackling various physics problems.
Distance-Time Relationship
The distance-time relationship is a fundamental concept with regards to motion. It describes how long it takes an object to cover a certain distance at a given speed. The key relationship is encapsulated in the formula
\[ t = \frac{d}{s} \]where:
  • \( t \) is time,
  • \( d \) is distance, and
  • \( s \) is speed.
This formula shows us that time is proportional to distance if the speed is constant. In the exercise, this relationship allowed us to calculate the time for each segment of the run.

To find average speed, we combine the distances and total time derived from each segment. This illustrates how distance and time interact to give us meaningful measures of motion. By understanding this relationship, one can better predict and analyze scenarios involving movement.

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Most popular questions from this chapter

Attempting to waste the last \(4.8 \mathrm{~s}\) of a basketball game, a player on the team that's ahead throws the ball straight up. He releases it \(1.6 \mathrm{~m}\) above the ground, and an opposing player catches it at the same height on the way down. The ball isn't permitted to touch the roof, \(17.2 \mathrm{~m}\) above the ground. Did the winning team run out the clock, or will the opponent have time to make a shot?

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