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(II) A horse canters away from its trainer in a straight line, moving \(116 \mathrm{~m}\) away in \(14.0 \mathrm{~s}\). It then turns abruptly and gallops halfway back in 4.8 s. Calculate \((a)\) its average speed and \((b)\) its average velocity for the entire trip, using "away from the trainer" as the positive direction.

Short Answer

Expert verified
The horse's average speed is approximately 9.26 m/s and its average velocity is approximately 3.09 m/s.

Step by step solution

01

Calculate Total Distance Traveled

First, we need to determine the total distance traveled by the horse. The horse moves 116 m away and then comes back halfway (which is half of 116 m). \[ \text{Total distance} = 116 \text{ m} + \frac{116}{2} \text{ m} = 174 \text{ m} \]
02

Calculate the Average Speed

Average speed is calculated using the total distance traveled divided by the total time taken. The total time is the sum of the time for each part of the journey: \(14.0 \text{ s} \) + \(4.8 \text{ s} \). \[ \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{174 \text{ m}}{18.8 \text{ s}} \approx 9.26 \text{ m/s} \]
03

Calculate the Total Displacement

Displacement is the straight-line distance from the start to the end point of the motion, taking into account direction. The horse ends halfway back to the starting point: \[ \text{Displacement} = 116 \text{ m} - 58 \text{ m} = 58 \text{ m} \]
04

Calculate the Average Velocity

Average velocity is calculated using the total displacement divided by the total time. \[ \text{Average velocity} = \frac{\text{Displacement}}{\text{Total time}} = \frac{58 \text{ m}}{18.8 \text{ s}} \approx 3.09 \text{ m/s} \]
05

Determine the Directions of Velocity and Speed

Since we set 'away from the trainer' as the positive direction, both distance and displacement calculations maintain this directionality. Average speed is independent of direction, while average velocity reflects movement direction, which is finally positive in this setup.

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Key Concepts

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

Average Velocity
In physics, average velocity is a vector quantity that describes the rate of change of displacement over time. Unlike average speed, average velocity considers direction. This makes it different because while speed is concerned with how fast something is moving, velocity tells us about both the speed and the direction of motion.

To calculate the average velocity, you use the formula:
  • Average Velocity = \( \frac{\text{Total Displacement}}{\text{Total Time}} \)
Because displacement is a vector, it is crucial to consider the starting and ending points, and not just how much ground was covered.

In the horse example, the horse's displacement was 58 meters after returning halfway. The total time for this journey was 18.8 seconds.
  • Thus, the average velocity is \( \frac{58 \text{ m}}{18.8 \text{ s}} \approx 3.09 \text{ m/s} \)
This calculation shows that the horse, on average, moved away from the trainer at 3.09 meters per second.
Displacement
Displacement refers to the straight-line distance between the starting point and the final position of an object, along with the direction of movement. It is a vector quantity, meaning it has both magnitude and direction, unlike distance which is a scalar quantity and only has magnitude.

In the context of the horse's movement, we consider 'away from the trainer' as the positive direction. The horse first moves 116 meters away from the trainer. Then, it returns halfway back, which is 58 meters, resulting in a net displacement of:
  • Displacement = 116 meters - 58 meters = 58 meters
Understanding displacement is essential for solving problems involving direction, as it provides true insight into an object's position change over a period of time.
Physics Problem Solving
Solving physics problems often involves breaking down a problem into smaller, manageable parts and using known formulas to find unknown values. This structured approach is essential in understanding complex physical phenomena, and it requires both analytical and critical thinking skills.

When you approach a physics problem like the horse scenario, consider:
  • Identifying given quantities and what you need to find.
  • Using appropriate formulas to link these quantities.
  • Being mindful of vector quantities and scalar quantities.
For the horse's motion:
  1. First, you determine the total distance traveled.
  2. Next, you calculate the displacement by considering direction.
  3. Third, you find average speed and average velocity.
Each step should logically lead to the next, making it easier to solve the problem accurately and efficiently. With practice, physics problem solving can become an intuitive and rewarding process.

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Most popular questions from this chapter

In putting, the force with which a golfer strikes a ball is planned so that the ball will stop within some small distance of the cup, say \(1.0 \mathrm{~m}\) long or short, in case the putt is missed. Accomplishing this from an uphill lie (that is, putting the ball downhill, see Fig. \(2-48\) ) is more difficult than from a downhill lie. To see why, assume that on a particular green the ball decelerates constantly at \(1.8 \mathrm{~m} / \mathrm{s}^{2}\) going downhill, and constantly at \(2.8 \mathrm{~m} / \mathrm{s}^{2}\) going uphill. Suppose we have an uphill lie \(7.0 \mathrm{~m}\) from the cup. Calculate the allowable range of initial velocities we may impart to the ball so that it stops in the range \(1.0 \mathrm{~m}\) short to \(1.0 \mathrm{~m}\) long of the cup. Do the same for a downhill lie \(7.0 \mathrm{~m}\) from the cup. What in your results suggests that the downhill putt is more difficult?

(I) At highway speeds, a particular automobile is capable of an acceleration of about \(1.8 \mathrm{~m} / \mathrm{s}^{2}\). At this rate, how long does it take to accelerate from \(80 \mathrm{~km} / \mathrm{h}\) to \(110 \mathrm{~km} / \mathrm{h} ?\)

(I) A stone is dropped from the top of a cliff. It is seen to hit the ground below after \(3.75 \mathrm{~s}\). How high is the cliff?

(I) What must your car's average speed be in order to travel \(235 \mathrm{~km}\) in \(3.25 \mathrm{~h} ?\)

A fugitive tries to hop on a freight train traveling at a constant speed of \(5.0 \mathrm{~m} / \mathrm{s}\). Just as an empty box car passes him, the fugitive starts from rest and accelerates at \(a=1.2 \mathrm{~m} / \mathrm{s}^{2}\) to his maximum speed of \(6.0 \mathrm{~m} / \mathrm{s} .(a)\) How long does it take him to catch up to the empty box car? (b) What is the distance traveled to reach the box car?

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