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(I) A particle at \(t_{1}=-2.0 \mathrm{~s}\) is at \(x_{1}=4.3 \mathrm{~cm}\) and at \(t_{2}=4.5 \mathrm{~s}\) is at \(x_{2}=8.5 \mathrm{~cm} .\) What is its average velocity? Can you calculate its average speed from these data?

Short Answer

Expert verified
The average velocity is approximately 0.646 cm/s. Average speed cannot be determined with the given data.

Step by step solution

01

Identify Given Data

First, note the values given in the problem: \(t_1 = -2.0 \; \mathrm{s}\), \(x_1 = 4.3 \; \mathrm{cm}\), \(t_2 = 4.5 \; \mathrm{s}\), and \(x_2 = 8.5 \; \mathrm{cm}\). We need to find the particle's average velocity.
02

Write Formula for Average Velocity

The formula for average velocity \(v_{avg}\) is given by \(v_{avg} = \frac{x_2 - x_1}{t_2 - t_1}\). It represents the change in position over the change in time.
03

Calculate Change in Position and Time

Calculate the change in position: \(x_2 - x_1 = 8.5 \; \mathrm{cm} - 4.3 \; \mathrm{cm} = 4.2 \; \mathrm{cm}\). Calculate the change in time: \(t_2 - t_1 = 4.5 \; \mathrm{s} - (-2.0 \; \mathrm{s}) = 6.5 \; \mathrm{s}\).
04

Calculate Average Velocity

Substitute the change in position and change in time into the average velocity formula: \[ v_{avg} = \frac{4.2 \; \mathrm{cm}}{6.5 \; \mathrm{s}} \approx 0.646 \; \mathrm{cm/s} \] This is the average velocity.
05

Consider Average Speed

The average speed is the total distance traveled divided by total time. However, we are only given two positions, not the path taken, so we cannot calculate the average speed from the provided data. It requires knowledge of the exact path or total distance.

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Key Concepts

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

Understanding Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. It acts as the foundation for understanding the behavior of objects in motion. In our exercise, we look into a particle moving along a straight line between two points in time, experiencing changes in its position.
By studying kinematics, students learn how variables such as time, position, and velocity interconnect. In problems like the one at hand, we utilize basic kinematic equations to describe motion accurately. These equations help us extract useful information from the changes observed, providing invaluable tools for understanding various physical scenarios.
Velocity Calculation Explained
Velocity is a vector quantity, which means it possesses both magnitude and direction. When calculating average velocity, we focus on the change in position divided by the change in time, as seen in the given formula: \[ v_{avg} = \frac{x_2 - x_1}{t_2 - t_1} \]This calculation measures how quickly an object changes its position from one place to another in a specific time interval.
  • \(x_1\) and \(x_2\) are the initial and final positions, respectively.
  • \(t_1\) and \(t_2\) are the initial and final times.
By substituting the given values, we obtain the rate of change in position, interpreted as average velocity. It is important to note the direction and role of time in ensuring signs align correctly with physical reality, especially when working with negative times. Students must also remember that velocity differs from speed because velocity includes direction.
Comprehending Average Speed
Average speed differs from average velocity as it considers only the magnitude of motion, disregarding direction. Essentially, it's a scalar quantity representing the total distance traveled divided by the total time: \[ \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} \]However, calculating average speed requires knowledge of the total distance covered, irrespective of the path taken.
In the exercise at hand, only the initial and final positions are provided, without details about the particle's path. Therefore, while average velocity could be calculated (using the straight-line distance between two points), the calculation of average speed requires more data about the journey's actual path. Understanding this distinction helps underline a profound aspect of motion analysis – the difference between the paths taken and just endpoint intervals.

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Most popular questions from this chapter

(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.

(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 \(110 \mathrm{~km} / \mathrm{h} ?\)

In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between stops. The more stops there are, the slower the train's average speed. To get an idea of this problem, calculate the time it takes a train to make a \(9.0-\mathrm{km}\) trip in two situations: \((a)\) the stations at which the trains must stop are \(1.8 \mathrm{~km}\) apart (a total of 6 stations, including those at the ends); and (b) the stations are \(3.0 \mathrm{~km}\) apart (4 stations total). Assume that at each station the train accelerates at a rate of \(1.1 \mathrm{~m} / \mathrm{s}^{2}\) until it reaches \(95 \mathrm{~km} / \mathrm{h},\) then stays at this speed until its brakes are applied for arrival at the next station, at which time it decelerates at \(-2.0 \mathrm{~m} / \mathrm{s}^{2} .\) Assume it stops at each intermediate station for \(22 \mathrm{~s}\).

(II) A car moving in a straight line starts at \(x=0\) at \(t=0\). It passes the point \(x=25.0 \mathrm{~m}\) with a speed of \(11.0 \mathrm{~m} / \mathrm{s}\) at \(t=3.00 \mathrm{~s} .\) It passes the point \(x=385 \mathrm{~m}\) with a speed of \(45.0 \mathrm{~m} / \mathrm{s}\) at \(t=20.0 \mathrm{~s}\). Find \((a)\) the average velocity and (b) the average acceleration between \(t=3.00 \mathrm{~s}\) and \(t=20.0 \mathrm{~s}\)

(II) A stone is thrown vertically upward with a speed of \(24.0 \mathrm{~m} / \mathrm{s}\). \((a)\) How fast is it moving when it reaches a height of \(13.0 \mathrm{~m} ?(b)\) How much time is required to reach this height? (c) Why are there two answers to (b)?

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