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Chapter 1: Problem 3

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 did 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 \(v\) : $$ t=\frac{l}{v}=\frac{17.0 \mathrm{~m}}{4.25 \mathrm{~m} / \mathrm{s}}=4.00 \mathrm{~s} $$

Short Answer

Expert verified
The journey took 4.00 seconds.

Step by step solution

01

Identify the given values

The problem provides a constant speed of \(4.25 \text{ m/s}\) and a distance of \(17.0 \text{ m}\).
02

Recall the formula for constant speed

The formula for constant speed is given by \(v = \frac{l}{t}\), where \(v\) is the speed, \(l\) is the distance, and \(t\) is the time taken to cover that distance.
03

Rearrange the formula to solve for time

Since we need to find the time \(t\), we rearrange the formula to make \(t\) the subject: \(t = \frac{l}{v}\).
04

Substitute values into the formula

Substitute the given values into the rearranged formula: \(t = \frac{17.0 \text{ m}}{4.25 \text{ m/s}}\).
05

Calculate the time taken

Perform the division: \(t = \frac{17.0}{4.25} = 4.00 \text{ s}\).

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Key Concepts

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

Distance Calculation
Distance is a fundamental concept in physics and everyday life. It is simply the measurement of how far an object has traveled from one point to another. Let's break down how it typically works:
  • Distance is usually measured in meters (m), kilometers (km), or miles in most real-world applications.
  • When an object moves at a constant speed, calculating the distance becomes straightforward. It can be visualized as the path between two points.
  • In the context of our problem, a robot travels a distance of 17.0 meters at a constant speed.
Understanding distance helps us comprehend movement and how it relates to time and speed. In any formula involving motion, distance forms a crucial part, allowing us to evaluate how much ground has been covered during a journey.
Time Calculation
Time is another crucial element in the study of motion. In this context, it refers to how long it takes to travel a certain distance at a specified speed. Here's how you can think about it:
  • Time is generally measured in seconds (s), minutes (min), or hours (h), depending on the scenario.
  • Using the formula for velocity, we can rearrange it to calculate the time: \[ t = \frac{l}{v} \tag{re-arranged formula} \]
  • For the robot's journey, by inserting the given values鈥攄istance (17.0 m) and speed (4.25 m/s)鈥攊nto the formula, we find: \[ t = \frac{17.0 \text{ m}}{4.25 \text{ m/s}} = 4.00 \text{ s} \]
This shows that it took the robot 4.00 seconds to cover the 17.0 meter distance. Time measurement is essential in determining how quick or slow a journey is, which is essential in planning and making predictions about travel.
Velocity Formula
The velocity formula is key to understanding how objects move. It captures the relationship between distance, time, and speed (or velocity):
  • The basic velocity formula is given by \[ v = \frac{l}{t} \]
  • Where \(v\) stands for velocity, \(l\) represents the distance covered, and \(t\) is the time taken.
  • In constant speed scenarios, velocity remains the same, making the calculations simpler.
By rearranging \[ v = \frac{l}{t} \] to \[ t = \frac{l}{v} \], it becomes easy to determine the time of a journey, as demonstrated in our exercise example. The velocity formula is pivotal in physics, not only because it helps in solving problems regarding motion but also because it provides a deeper understanding of how different factors like speed, time, and distance interplay during movement.

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Most popular questions from this chapter

A kid stands \(6.00 \mathrm{~m}\) from the base of a flagpole which is \(8.00 \mathrm{~m}\) tall. Determine the magnitude of the displacement of the brass eagle on top of the pole with respect to the youngster's feet. The geometry corresponds to a 3-4-5 right triangle (i.e., \(3 \times 2-4 \times 2-5 \times 2\) ). Thus, the hypotenuse, which is the 5 -side, must be \(10.0 \mathrm{~m}\) long, and that's the magnitude of the displacement.

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. 1-10. 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.

Starting at the origin of coordinates, the following displacements are made in the \(x y\) -plane (that is, the displacements are coplanar): \(60 \mathrm{~mm}\) in the \(+y\) -direction, \(30 \mathrm{~mm}\) in the \(-x\) -direction, \(40 \mathrm{~mm}\) at \(150^{\circ}\), and \(50 \mathrm{~mm}\) at \(240^{\circ}\). Find the resultant displacement both graphically and algebraically.

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.

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.
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