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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: $$ v_{a v}=\frac{l}{t}=\frac{10.0 \times 10^{3} \mathrm{~m}}{1800 \mathrm{~s}}=5.56 \mathrm{~m} / \mathrm{s} $$

Step by step solution

01

Convert Distance to Meters

The given distance is 10.0 km. We need to convert this distance into meters using the conversion factor: 1 km = 1000 m. Thus, \(10.0 \text{ km} \times 1000 \text{ m/km} = 10,000 \text{ m}\).

02

Convert Time to Seconds

The time taken for the travel is given as 30.0 minutes. To convert this into seconds, use the conversion factor: 1 minute = 60 seconds. Therefore, \(30.0 \text{ min} \times 60 \text{ s/min} = 1800 \text{ s}\).

03

Calculate the Average Speed

Using the equation for average speed \(v_{av} = \frac{l}{t}\), where \(l\) is the distance in meters and \(t\) is the time in seconds, we plug in the values: \(v_{av} = \frac{10,000 \text{ m}}{1800 \text{ s}} = 5.56 \text{ m/s}\).

04

Express with Correct Significant Figures

Since both the distance and time are provided with three significant figures, the calculated average speed \(5.56 \text{ m/s}\) is also expressed with three significant figures.

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Key Concepts

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

Unit Conversion

Converting units is an essential skill in solving physics problems. It involves changing measurements from one unit to another using conversion factors.
In our exercise, we start by converting kilometers to meters and minutes to seconds. This is because the standard unit of speed is meters per second (m/s).

• Kilometers to Meters: Since 1 kilometer equals 1000 meters, you multiply the given distance (10.0 km) by 1000 to convert it to meters, resulting in 10,000 meters.
• Minutes to Seconds: Since 1 minute equals 60 seconds, multiply the given time (30.0 minutes) by 60 to convert it to seconds, which gives you 1800 seconds.

Why convert? Converting ensures that your calculations comply with standard scientific units, making the results both valid and universally comparable.

Significant Figures

When performing calculations in physics, it is crucial to maintain accuracy through the use of significant figures. Significant figures refer to the digits in a number that contribute to its precision.
Both measured and calculated values must be expressed correctly.

• Preserving Precision: In our exercise, the measurements for distance (10.0 km) and time (30.0 min) are given with three significant figures. Thus, the calculated average speed must also reflect this level of precision: 5.56 m/s.
• Importance: Calculating with the correct number of significant figures ensures that results are not over or under-represented in terms of data accuracy. It maintains the integrity of the calculation's limitations.

Always round off your final answer to the proper number of significant figures based on your least precise measurement in the problem.

Kinematics

Kinematics is the study of motion without considering the forces causing it. In this context, we're interested in calculating the average speed of the car in our problem.
Average speed is given by an equation derived from kinematics:

• Formula: The formula for average speed is \( v_{av} = \frac{l}{t} \) where \( l \) is the distance traveled and \( t \) is the time taken to travel that distance.
• Application: This formula allows us to understand how far an object moves per unit of time, giving a clear picture of its motion across a specific distance and timeframe.

Kinematics helps students to link time, distance, and speed effectively, developing a deeper understanding of how these aspects of motion are interconnected.

Distance and Time

Understanding the relationship between distance and time is fundamental in physics, especially when calculating speed. Distance and time are two of the main variables used in kinematic equations to describe motion.

• Distance: It refers to the total length of the path taken during motion. In this exercise, the distance is given as 10.0 km, which we convert to 10,000 meters for standard calculation.
• Time: It is the duration over which the motion occurs. Here the time is 30.0 minutes, converted to 1800 seconds.

These variables, when used together in the speed equation, help determine how quickly or slowly an object is moving. Analyzing everyday scenarios like a drive to school reveals the practical applications of these foundational concepts.