/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 A runner makes one complete lap ... [FREE SOLUTION] | 91Ó°ÊÓ

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A runner makes one complete lap around a 200 -m track in a time of \(25 \mathrm{~s}\). What were the runner's \((a)\) average speed and \((b)\) average velocity? (a) From the definition, $$ \text { Average speed }=\frac{\text { Distance traveled }}{\text { Time taken }}=\frac{200 \mathrm{~m}}{25 \mathrm{~s}}=8.0 \mathrm{~m} / \mathrm{s} $$ (b) Because the run ended at the starting point, the displacement vector from starting point to end point has zero length. Since \(\overrightarrow{\mathbf{v}}_{w v}=\overrightarrow{\mathbf{s}} / t\) $$ \left|\overrightarrow{\mathbf{v}}_{a v}\right|=\frac{0 \mathrm{~m}}{25 \mathrm{~s}}=0 \mathrm{~m} / \mathrm{s} $$

Short Answer

Expert verified
Average speed = 8.0 m/s; Average velocity = 0 m/s.

Step by step solution

01

Understanding Average Speed

The average speed of a runner or any object describes the total distance traveled divided by the total time taken for the journey. It is calculated as:\[\text{Average speed} = \frac{\text{Distance traveled}}{\text{Time taken}}\].
02

Calculate Distance and Time

The runner completes a lap of 200 meters and it takes them 25 seconds. Therefore, we have:\[\text{Distance traveled} = 200 \text{ meters}, \text{Time taken} = 25 \text{ seconds}\].
03

Calculate Average Speed

Substitute the values of distance and time into the average speed formula:\[\text{Average speed} = \frac{200 \text{ meters}}{25 \text{ seconds}}\]This results in:\[\text{Average speed} = 8.0 \text{ m/s}\].
04

Understanding Average Velocity

Average velocity is defined as the total displacement divided by the total time taken. Displacement is the straight line distance between the starting and ending points of the journey.
05

Calculate Displacement and Velocity

Since the runner ends at the starting point after completing the lap, the displacement is zero; hence, \[\text{Displacement} = 0 \text{ m}\]. Therefore, the average velocity is:\[\text{Average velocity} = \frac{0 \text{ m}}{25 \text{ seconds}} = 0 \text{ m/s}\].

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Key Concepts

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

Distance and Displacement
When measuring motion, it's crucial to understand the difference between distance and displacement. Distance is the total path covered by an object, regardless of direction. In our runner's case, the total distance is 200 meters from start to finish, encircling the track. Displacement, however, is a vector quantity that measures the shortest straight line between the starting and ending points. If a runner starts and finishes at the same point, as in a circular track, the displacement is zero. This happens because the initial and final positions overlap, leading to no change in position. Here, even though the distance is 200 meters, the displacement is 0 meters because the starting and ending points are identical. Distinguishing between these two concepts is vital in physics, as they often yield different values when calculating speed and velocity.
Time Calculation
In solving physics problems, calculating time is a basic yet essential step. Here, time represents the duration of motion or how long an event continues. The runner's lap around the track lasted 25 seconds. Knowing the time taken allows us to determine rates such as speed and velocity. Time measurement is crucial for calculating the average speed as well as understanding the dynamics of movement. It often serves as the common denominator in kinematics, helping to relate differing motion aspects together. For instance, when given distance and time, we can easily determine speed. The calculated time can affect outcomes in calculations, particularly when comparing speeds or evaluating how changes in time alter motion dynamics. Thus, time calculation is simple but forms a backbone for understanding motion in physics.
Physics Problems
Physics problems frequently involve applying concepts like distance, displacement, and time to address real-world scenarios. These problems enhance comprehension by connecting theoretical knowledge to practical application. In our example, the challenge was to calculate average speed and average velocity. This required identifying the runner's path length and displacement. By distinguishing between these two measures, we could solve each part of the problem accurately. For the average speed, we used the entire path length over the given time, while for average velocity, we concentrated on the net change in position over the time. Through such problems, students practice analyzing different facets of motion and hone their ability to apply conceptual knowledge. Real-world problems encourage critical thinking, offering a chance to observe how physics principles manifest in everyday activities.

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Most popular questions from this chapter

A bug starts at point \(A\), crawls \(8.0 \mathrm{~cm}\) east, then \(5.0 \mathrm{~cm}\) south, \(3.0\) \(\mathrm{cm}\) west, and \(4.0 \mathrm{~cm}\) north to point \(B\). \((a)\) How far south and east is \(B\) from \(A\) ? \((b)\) Find the displacement from \(A\) to \(B\) both graphically and algebraically.

Find the \(x\) - and \(y\) -components of a 25.0-m displacement at an angle of \(210.0^{\circ}\) The vector displacement and its components are depicted in Fig. 1-11. The scalar components are $$ \begin{array}{l} x \text { -component }=-(25.0 \mathrm{~m}) \cos 30.0^{\circ}=-21.7 \mathrm{~m} \\\ y \text { -component }=-(25.0 \mathrm{~m}) \sin 30.0^{\circ}=-12.5 \mathrm{~m} \end{array} $$ Notice in particular that each component points in the negative coordinate direction and must therefore be taken as negative.

A model plane flew \(100 \mathrm{~m}\) in \(25.0 \mathrm{~s}\) followed by another \(240 \mathrm{~m}\) in an additional \(60.0 \mathrm{~s}\), whereupon it crashed into the ground. How far did it travel in total? How long was it in the air? What was its average speed? [Hint: The overall average is not equal to the average of the averages. When you have several segments in a problem, label them like this: \(l_{1}\) and \(l_{2}\) and \(t_{1}\) and \(t_{2}\), such that \(l=l_{1}\) \(+l_{2}\) and \(\left.t=t_{1}+t_{2} .\right]\)

Starting from the center of town, a car travels east for \(80.0 \mathrm{~km}\) and then turns due south for another \(192 \mathrm{~km}\), at which point it runs out of gas. Determine the displacement of the stopped car from the center of town.

Rolling along across the machine shop at a constant speed of \(4.25\) \(\mathrm{m} / \mathrm{s}\), a robot covers a distance of \(17.0 \mathrm{~m}\). How long does that journey take? Since the speed is constant the defining equation is \(v=l / t\). Multiply both sides of this expression by \(t\) and then divide both by \(u:\) $$ t=\frac{l}{v}=\frac{17.0 \mathrm{~m}}{4.25 \mathrm{~m} / \mathrm{s}}=4.00 \mathrm{~s} $$

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