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In an \(8,00 \mathrm{km}\) race, one runner runs at a steady \(11.0 \mathrm{km} / \mathrm{h}\) and another runs at \(14.0 \mathrm{km} / \mathrm{h}\). How far from the finish line is the slower runner when the faster runner finishes the race?

Short Answer

Expert verified
The slower runner is 1.73 km from the finish line when the faster runner finishes the race.

Step by step solution

01

Determine time taken by the faster runner to complete the race

We can calculate time using the formula for speed \(speed = distance/time\). We need to rearrange it to find time: \(time = distance/speed\). Substituting values: \(time = 8 km / 14 km/h = 0.57 hours\)
02

Calculate the distance covered by the slower runner in this time

We again use the same formula for speed, but this time we need to find distance: \(distance = speed * time\). Substituting values: \(distance = 11 km/h * 0.57 hours = 6.27 km\)
03

Calculate the remaining distance for the slower runner

The total distance of the race is 8 km and the slower runner has covered 6.27 km in the time it took for the faster runner to finish. Thus, the distance remaining for the slower runner is \(8 km - 6.27 km = 1.73 km\)

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Key Concepts

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

Speed and Velocity
Understanding the concepts of speed and velocity is essential in kinematics, which is the branch of physics concerned with the motion of objects. Speed is a scalar quantity, meaning it only has magnitude and no direction. It tells us how fast an object is moving. If a car is traveling at 60 kilometers per hour, that is its speed.

Distinguishing Speed from Velocity

Velocity, on the other hand, is a vector quantity; it has both magnitude and direction. For example, if we say a plane is moving northeast at 500 kilometers per hour, we’re describing its velocity. In physics problems, it’s crucial to know whether the question is asking for speed or velocity to provide the correct information.

When solving problems related to speed and velocity, we use the basic formula: \[ speed = \frac{distance}{time} \.\] However, it is important to remember that when dealing with velocity, direction is also part of the equation. This distinction plays a crucial role in problems involving relative motion, which we will explore in another section.
Distance-Time Calculation
The calculation of distance over time is fundamental to solving kinematics problems. By knowing the speed at which an object is traveling and the time it has been moving, we can calculate how far it has gone. The formula \[ distance = speed \times time \.\] is incredibly useful for predicting where an object will be at a certain point in its journey.

Putting it Into Practice

Let’s consider a practical example. Say a cyclist is traveling at a constant speed of 15 km/h, and we want to find out how far the cyclist will have ridden after 3 hours. The calculation would be straightforward: \[ distance = 15 km/h \times 3 h = 45 km \.\] This type of calculation is a staple in kinematics and is the backbone for more complex concepts like relative motion.
Relative Motion
The concept of relative motion is crucial when analyzing scenarios where two or more objects are in motion at the same time. It takes into account not only the speed of each object but also their directions and how these affect their positions relative to each other over time.

Relative Motion in Action

For instance, if we have two trains traveling in the same direction on parallel tracks, one faster than the other, the faster train will pass the slower one. The question of how far apart they are at any given time depends on their relative speeds - the difference in their speeds, to be precise. If Train A is going 20 km/h faster than Train B, Train A will increase the distance between them by 20 kilometers each hour they travel.

Using relative motion helps us understand the dynamics between objects, such as the slower and faster runners in the race example. By comprehending their relative speeds, we can determine the precise moment or location when they interact or relate a certain way relative to the finish line or starting point.

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Most popular questions from this chapter

A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.0 s later. What was the rocket's acceleration?

When jumping, a flea reaches a takeoff speed of \(1.0 \mathrm{m} / \mathrm{s}\) over a distance of \(0.50 \mathrm{mm}\) a. What is the flea's acceleration during the jump phase? b. How long does the acceleration phase last? c. If the flea jumps straight up, how high will it go? (Ignore air resistance for this problem; in reality, air resistance plays a large role, and the flea will not reach this height.)

A Porsche challenges a Honda to a \(400 \mathrm{m}\) race. Because the Porsche's acceleration of \(3.5 \mathrm{m} / \mathrm{s}^{2}\) is larger than the Honda's \(3.0 \mathrm{m} / \mathrm{s}^{2},\) the Honda gets a \(100-\mathrm{m}\) head start-it is only \(300 \mathrm{m}\) from the finish line. Assume, somewhat unrealistically, that both cars can maintain these accelerations the entire distance. Who wins, and by how much time?

In a \(5.00 \mathrm{km}\) race, one runner runs at a steady \(12.0 \mathrm{km} / \mathrm{h}\) and another runs at \(14.5 \mathrm{km} / \mathrm{h}\). How long does the faster runner have to wait at the finish line to see the slower runner cross?

While running a marathon, a long-distance runner uses a stopwatch to time herself over a distance of \(100 \mathrm{m}\). She finds that she runs this distance in 18 s. Answer the following by considering ratios, without computing her velocity. a. If she maintains her speed, how much time will it take her to run the next \(400 \mathrm{m} ?\) b. How long will it take her to run a mile at this speed?

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