Question

Driving to School The time \(t\) that it takes to get to school varies inversely with your average speed \(s\). (a) Suppose that it takes you 40 minutes to get to school when your average speed is 30 miles per hour. Express the driving time to school in terms of average speed. (b) Suppose that your average speed to school is 40 miles per hour. How long will it take you to get to school?

Short answer

a) \( t = \frac{1200}{s} \) b) 30 minutes

Step-by-step solution

Understand Inverse Variation

Since time \(t\) varies inversely with speed \(s\), it can be expressed as \(t = \frac{k}{s}\) where \(k\) is a constant.

Find the Constant \(k\)

Use the given values to find \(k\). When \(t = 40\) minutes and \(s = 30\) miles per hour, substitute into the equation: \[ \ 40 = \frac{k}{30}\] Multiply both sides by 30: \[ \ k = 40 \times 30 = 1200 \]

Express Driving Time in Terms of Speed

Now that we know \(k\), we can rewrite the equation: \[ \ t = \frac{1200}{s} \] This is the expression for driving time in terms of average speed.

Substitute Speed to Find Time

Use the expression derived in the previous step to find the time it takes to get to school when the speed is 40 miles per hour: \[ t = \frac{1200}{40} \]

Calculate the Time

Perform the division: \[ t = \frac {1200}{40} = 30 \] Thus, it will take 30 minutes to get to school.

Key concepts

Inverse Variation

Inverse variation is a relationship between two variables where one variable increases while the other decreases. In this context, we are considering the relationship between the time it takes to get to school and the average speed.
When one variable (time) decreases, the other variable (speed) must increase to maintain the same distance. This can be mathematically represented as: \[ t = \frac{ k } { s } \]
Here, \( t \) is the time, \( s \) is the speed, and \( k \) is the constant of variation. This equation shows that as speed increases, the time taken decreases proportionately.
Let's break this down with an example. If it takes you 40 minutes to get to school at a speed of 30 miles per hour, you can find this constant \( k \) by multiplying time and speed:
\[ 40 \times 30 = 1200 \]
So, \( k = 1200 \). This constant can now be used to express the relationship for any speed or time.

Average Speed Calculation

Average speed is how fast you are traveling on average over a certain distance or duration. To calculate average speed, you typically use the formula:
\[ \text{Average Speed} = \frac{ \text{Total Distance} } { \text{Total Time} } \]
However, with inverse variation in the given problem, we can find the average speed needed to maintain the calculated time. For example, when your speed is 40 miles per hour, you use the inverse equation:
\[ t = \frac{ k } { s } \]
Here, by substituting \( k = 1200 \) and \( s = 40 \) miles per hour, we can calculate the time taken as:
\[ t = \frac{ 1200 } { 40 } = 30 \text{ minutes} \]
This calculation shows that at higher speeds, the time to travel decreases, illustrating the inverse relationship.

Distance-Time Relationship

Understanding the distance-time relationship is crucial in solving travel-related problems. This relationship is fundamentally grounded on the formula:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]
In the context of traveling to school, if your speed and time are inversely related, adjusting one will directly affect the other. For example, sticking with our given problem, if it takes 40 minutes at 30 miles per hour, keeping the distance constant ensures:
\[ \text{Distance} = 30 \times \frac{ 40 } { 60 } = 20 \text{ miles} \]
This confirms that regardless of speed, as long as the product of speed and time matches this constant distance, the relationship holds. So if speeds change, time compensates.

Constant of Variation

The constant of variation \( k \) plays a key role in inverse variation problems. This constant remains unchanged despite variations in speed or time. In our problem, we found the constant \( k \) using:
\[ k = \text{Speed} \times \text{Time} \]
Substituting given values, we have:
\( k = 30 \times 40 = 1200 \)
This constant is pivotal because it lets you express time for any given speed:
\[ t = \frac{ 1200 } { s } \]
For instance, should the speed rise to 50 miles per hour, using our formula gives:
\( t = \frac{ 1200 } { 50 } = 24 \text{ minutes} \)
This constant underlies the entire problem, enabling us to convert and solve different speed-time scenarios while keeping the distance traveled consistent.