/*! 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 1 A toy train moves along a windin... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 1

A toy train moves along a winding track at an average speed of \(0.25 \mathrm{~m} / \mathrm{s}\). How far will it travel in \(4.00\) minutes? The defining equation is \(v_{a v}=l / t\). Here \(l\) is in meters, and \(t\) is in seconds, so the first thing to do is convert \(4.00\) min into seconds: \((4.00 \min )(60.0 \mathrm{~s} / \mathrm{min})=240 \mathrm{~s}\). Solving the equation for \(l\). $$ l=v_{a v} t=(0.25 \mathrm{~m} / \mathrm{s})(240 \mathrm{~s}) $$ Since the speed has only two significant figures, \(l=60 \mathrm{~m}\).

Short Answer

Expert verified
The toy train will travel 60 meters in 4 minutes.

Step by step solution

01

Convert Time from Minutes to Seconds

The problem gives the time in minutes, but we need it in seconds for our equation. Since there are 60 seconds in a minute, multiply the time in minutes by 60: \( 4.00 \text{ minutes} \times 60 \text{ seconds/minute} = 240 \text{ seconds} \).
02

Use the Average Speed Formula

The formula for distance \( l \) when speed \( v \) and time \( t \) are known is \( l = v \times t \). Substitute the known values: \( v = 0.25 \text{ m/s} \) and \( t = 240 \text{ seconds} \).
03

Calculate the Distance

Multiply the average speed by the time to find the distance: \( l = 0.25 \text{ m/s} \times 240 \text{ s} = 60 \text{ m} \).
04

Consider Significant Figures

The given speed has two significant figures (0.25), so the final answer should also have two significant figures. Thus, \( l = 60 \text{ m} \).

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Key Concepts

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

Unit Conversion
When working on physics problems, unit conversion is a critical step that ensures all calculations are performed in a consistent unit system. For example, in the scenario of a toy train moving at an average speed, it is essential to convert the time from minutes to seconds because the speed is given in meters per second (m/s). To convert minutes to seconds, remember:
  • There are 60 seconds in a minute.
  • Multiply the number of minutes by 60 to get the total seconds.
For instance, if the train runs for 4.00 minutes, multiply 4.00 by 60 to get 240 seconds. This consistency in units allows you to use the speed formula effectively, without risking errors due to incompatible units.
Significant Figures
Significant figures play an essential role in scientific calculations, as they convey the precision of the measured values. When dealing with multiplication or division, the number of significant figures in the result should match the number in the least precise measurement from the numbers involved.
  • In our example, the speed has two significant figures: 0.25 m/s.
  • The calculated time (240 s) is precise to three figures, but the final result (distance) should reflect the precision of the least precise measurement.
Thus, the calculated distance should be reported as 60 meters, using two significant figures to maintain consistency with the speed's precision.
Distance Calculation
Distance calculation using speed and time is a straightforward process if the units are consistent. The formula to calculate distance is:\[ l = v \times t \]where:
  • \( l \) is the distance.
  • \( v \) is the average speed.
  • \( t \) is the time.
To find the distance the toy train travels in the given time, substitute the average speed and converted time into the equation. For instance, with a speed of 0.25 m/s and a time of 240 seconds, the distance is:\[ l = 0.25 \times 240 = 60 \text{ meters} \]This straightforward multiplication gives you the total distance traveled, ensuring all units are accurately incorporated into the calculation.

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Most popular questions from this chapter

A ship is traveling due east at \(10 \mathrm{~km} / \mathrm{h}\). What must be the speed of a second ship heading \(30^{\circ}\) east of north if it is always due north of the first ship?

Compute algebraically the resultant of the following coplanar displacements: \(20.0 \mathrm{~m}\) at \(30.0^{\circ}, 40.0 \mathrm{~m}\) at \(120.0^{\circ}\) \(25.0 \mathrm{~m}\) at \(180.0^{\circ}, 42.0 \mathrm{~m}\) at \(270.0^{\circ}\), and \(12.0 \mathrm{~m}\) at \(315.0^{\circ} .\) Check your answer with a graphical solution.

Determine the displacement vector that must be added to the displacement \((25 \hat{\mathbf{i}}-16 \hat{\mathbf{j}}) \mathrm{m}\) to give a displacement of \(7.0 \mathrm{~m}\) pointing in the \(+x\) -direction?

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.

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