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Chapter 2: Problem 3

(I) A particle at \(t_1 = -\)2.0 s is at \(x_1 =\) 4.8 cm and at \(t_2 =\) 4.5 s is at \(x_2 =\) 8.5 cm. What is its average velocity over this time interval? Can you calculate its average speed from these data?Why or why not?

Short Answer

Expert verified
The average velocity is approximately 0.569 cm/s. We cannot calculate the average speed with the given data.

Step by step solution

01

Understand the Problem

We need to calculate the average velocity of a particle given its initial and final positions and times. The initial time \( t_1 \) is \(-2.0\,s\) with position \( x_1 = 4.8\,cm \), and the final time \( t_2 \) is \( 4.5\,s \) with position \( x_2 = 8.5\,cm \).
02

Use the Formula for Average Velocity

The average velocity \( v_{avg} \) is calculated using the formula:\[v_{avg} = \frac{x_2 - x_1}{t_2 - t_1}\]where \( x_2 \) and \( x_1 \) are the final and initial positions, and \( t_2 \) and \( t_1 \) are the final and initial times.
03

Plug in the Values and Solve

Substitute the given values:\[v_{avg} = \frac{8.5\,cm - 4.8\,cm}{4.5\,s - (-2.0\,s)} = \frac{3.7\,cm}{6.5\,s} \approx 0.569\,cm/s\]The average velocity of the particle is approximately \( 0.569\,cm/s \).
04

Determine if Average Speed Can Be Calculated

Average speed is the total distance traveled divided by the total time taken. To calculate it, we need to know the exact path the particle took between the two points. Since the data only provides initial and final positions, we cannot determine if the path was straight or if there were deviations. Therefore, we cannot calculate the average speed without additional information.

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that focuses on the motion of objects without regard to the causes of this motion. It is all about understanding and describing how things move. In kinematics, we primarily deal with quantities such as displacement, velocity, and acceleration.
These quantities help us to capture the essence of how something moves from one point to another over time. For example, in the given exercise, kinematics is employed to determine how fast a particle is moving by calculating the average velocity between two time intervals. Some key points about kinematics include:
  • It helps in determining how a particle travels from one location to another.
  • The focus is on quantities like displacement (change in position), velocity, and time intervals.
  • Kinematics does not concern itself with why or how the object started moving, just the nature and specifics of the movement.
Average Speed
Average speed is an important concept in kinematics, distinct from average velocity. While average velocity considers the direction of travel, average speed does not. It looks purely at how fast an object is moving along its path. The calculation for average speed can be summarized as: - Total distance traveled divided by the total time of travel. It's important to note, as in the original exercise, average speed can only be calculated if the exact path is known.

Even if the particle ends up at a point further from its start, its path could have involved turns or even retracing its path, affecting the total distance.
  • Average speed equals total path length divided by total time interval.
  • It is always a positive quantity because it has no direction.
  • Cannot be calculated with only initial and final position data, unless the path is known.
Position and Displacement
Understanding the difference between position and displacement is crucial in kinematics. Position refers to where an object is located at a specific time, usually described using coordinates. In the exercise, the particle's positions are given as 4.8 cm and 8.5 cm. Displacement, on the other hand, is the change in position, and it takes into account the direction of movement. It is a vector quantity, which means it has both magnitude and direction.

In comparison to distance, which is scalar, displacement tells us how far out of place an object is relative to its starting point.
  • Position is simply the location of the object at any given time.
  • Displacement is calculated as the final position minus the initial position.
  • Considered a vector quantity, encompassing both magnitude and direction.

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Most popular questions from this chapter

(I) A bird can fly 25 km/h. How long does it take to fly 3.5 km?

(II) A car traveling at 95 km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80 m. What was the magnitude of the average acceleration of the driver during the collision? Express the answer in terms of "\(g\)'s," where 1.00 \(g =\) 9.80 m/s\(^2\).

In the design of a \(\textbf{rapid transit system}\), it is necessary to balance the average speed of a train against the distance between station 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 15.0-km trip in two situations: (\(a\)) the stations at which the trains must stop are 3.0 km apart (a total of 6 stations, including those at the ends); and (\(b\)) the stations are 5.0 km apart (4 stations total). Assume that at each station the train accelerates at a rate of 1.1 m/s\(^2\) until it reaches 95 km/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 m/s\(^2\). Assume it stops at each intermediate station for 22 s.

A race car driver must average 200.0 km/h over the course of a time trial lasting ten laps. If the first nine laps were done at an average speed of 196.0 km/h, what average speed must be maintained for the last lap?

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