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Chapter 1: Problem 19

An ant walked \(10.0 \mathrm{~cm}\) across the floor in \(6.2 \mathrm{~s}\). What was its average speed in \(\mathrm{m} / \mathrm{s}\) ? [Hint: 2 significant figures. You are given \(s\) and \(t\) and must find uau. Watch out for units.]

Short Answer

Expert verified
The average speed is approximately \( 0.016 \text{ m/s} \).

Step by step solution

01

Identify the Given Variables

We know the distance \( d \) the ant traveled is \( 10.0 \text{ cm} \) and the time \( t \) it took is \( 6.2 \text{ s} \).
02

Convert Units

Since we need the speed in \( \text{m/s} \) but the distance is in centimeters, we convert \( 10.0 \text{ cm} \) to meters. We know that 1 cm = 0.01 m, so \( 10.0 \text{ cm} = 10.0 \times 0.01 = 0.10 \text{ m} \).
03

Formula for Average Speed

Recall the formula for average speed \( v \): \[ v = \frac{d}{t} \] where \( d \) is the distance in meters and \( t \) is the time in seconds.
04

Insert Values and Calculate

Using the formula \( v = \frac{d}{t} \), substitute \( d = 0.10 \text{ m} \) and \( t = 6.2 \text{ s} \): \[ v = \frac{0.10}{6.2} \approx 0.0161 \text{ m/s} \].
05

Round to Significant Figures

The problem specifies to use two significant figures. Thus, we round \( 0.0161 \text{ m/s} \) to two significant figures, resulting in: \( 0.016 \text{ m/s} \).

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Key Concepts

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

Distance Conversion
When calculating average speed, it's important to convert all units to the same standards. In our problem, the ant's journey is measured in centimeters, but for calculating speed in meters per second, we must convert this distance to meters.

To convert centimeters to meters, remember that 1 centimeter is equivalent to 0.01 meters. Thus, to convert 10.0 cm to meters, we multiply by 0.01.
  • Given: 10.0 cm
  • Conversion factor: 1 cm = 0.01 m
Therefore, the conversion is calculated as follows:
\[ 10.0 \text{ cm} \times 0.01 = 0.10 \text{ m} \]
This step is essential for uniformity in further calculations and is a basic step in physics and chemistry whenever different units are involved.
Significant Figures
Significant figures are critical in scientific calculations because they indicate the precision of a measurement. Calculations should be rounded to reflect the accuracy of the recorded data.

In our problem, we are required to report the average speed using two significant figures. Significant figures are usually derived from the given data which, in this case, are 10.0 cm and 6.2 s.

Steps to identify significant figures:
  • All non-zero digits are significant.
  • Zeros between non-zero digits are significant.
  • Trailing zeros in a decimal number are significant.
In the value 0.0161 m/s, all these apply. To focus on two significant figures, 0.0161 is approximated to 0.016 m/s.
Speed Calculation
To determine the average speed of the ant, we utilize the formula that defines speed as distance traveled over time taken.

The formula is:
\[ v = \frac{d}{t} \]
In this context:
  • \(d = 0.10\text{ m} \)
  • \( t = 6.2 \text{ s} \)
Inserting these values into the formula gives the average speed: \[ v = \frac{0.10}{6.2} \approx 0.0161 \text{ m/s} \].
After calculating, it's important to check for the proper significant figures, leading us to round the speed to 0.016 m/s.
This calculation method is essential for determining consistent and comparable speed results.

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Most popular questions from this chapter

A little turtle is placed at the origin of an \(x y\) -grid drawn on a large sheet of paper. Each grid box is \(1.0 \mathrm{~cm}\) by \(1.0 \mathrm{~cm}\). The turtle walks around for a while and finally ends up at point \((24,10)\), that is, 24 boxes along the \(x\) -axis, and 10 boxes along the \(y\) -axis. Determine the displacement of the turtle from the origin at the point.

Find the scalar \(x\) - and \(y\) -components of the following displacements in the \(x y\) -plane: (a) \(300 \mathrm{~cm}\) at \(127^{\circ}\) and (b) \(500 \mathrm{~cm}\) at \(220^{\circ}\).

A runner travels \(1.5\) laps around a circular track in a time of \(50 \mathrm{~s}\). The diameter of the track is \(40 \mathrm{~m}\) and its circumference is \(126 \mathrm{~m}\). Find \((a)\) the average speed of the runner and \((b)\) the magnitude of the runner's average velocity. Be careful here; average speed depends on the total distance traveled, whereas average velocity depends on the displacement at the end of the particular journey.

A boat, propelled so as to travel with a speed of \(0.50 \mathrm{~m} / \mathrm{s}\) in still water, moves directly across a river that is \(60 \mathrm{~m}\) wide. The river flows with a speed of \(0.30 \mathrm{~m} / \mathrm{s}\). (a) At what angle, relative to the straight-across direction, must the boat be pointed? (b) How long does it take the boat to cross the river?

A student driving a car travels \(10.0 \mathrm{~km}\) in \(30.0\) min. What was her average speed? The defining equation is \(u_{a v}=l / t\). Here \(l\) is in kilometers, and \(t\) is in minutes, so the first thing to do is convert \(10.0 \mathrm{~km}\) to meters and then \(30.0\) min into seconds: \((10.0 \mathrm{~km})(1000 \mathrm{~m} / \mathrm{km})=10.0 \times\) \(10^{3} \mathrm{~m}\) and \((30.0 \mathrm{~min}) \times(60.0 \mathrm{~s} / \mathrm{min})=1800 \mathrm{~s}\). We need to solve for \(v_{a u}\), giving the numerical answer to three significant figures: $$ v_{a v}=\frac{l}{t}=\frac{10.0 \times 10^{3} \mathrm{~m}}{1800 \mathrm{~s}}=5.56 \mathrm{~m} / \mathrm{s} $$
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