Problem 26 - Contemporary Precalculus

Question

To get to work, Sam jogs 3 kilometers to the train and then rides the remaining 5 kilometers. If the train goes 40 kilometers per hour faster than Sam's constant rate of jogging and the entire trip takes 30 minutes, how fast does Sam jog?

Step-by-Step Solution

Answer

Answer: Sam's jogging speed is 6 km/h.

Step 1: Write down the given information

We are given the following information: Sam jogs 3 kilometers to the train, rides the train for 5 kilometers, the total trip takes 30 minutes (0.5 hours), and the train is 40 km/h faster than Sam's jogging speed. Let's denote Sam's jogging speed as S km/h and the train's speed as T km/h. Thus, we have: S + 40 km/h = T.

Step 2: Set up relationships for speed, distance, and time for Sam's jogging and the train ride

We will use the relation speed = distance/time to set up an equation for Sam's jogging and the train ride. We know the total time is 0.5 hours. Let's denote the time Sam spends jogging as J hours and the time spent on the train as R hours. For Sam's jogging: S = 3/J For the train ride: T = 5/R

Step 3: Set up an equation for the total time

The total time for the trip is J hours (jogging) + R hours (train ride) = 0.5 hours. So, we have: J + R = 0.5

Step 4: Solve for J and R in terms of S and T

We can get J and R in terms of S and T by using the equations for Sam's jogging and the train ride: J = 3/S R = 5/T

Step 5: Substitute the expressions for J and R in terms of S and T into the total time equation

Now, we can substitute the expressions for J and R into the total time equation: (3/S) + (5/T) = 0.5

Step 6: Substitute for T using the speed relationship equation

Since we have the relationship: S + 40 = T, we can substitute this into the total time equation: (3/S) + (5/(S+40)) = 0.5

Step 7: Solve for Sam's jogging speed (S)

To solve for Sam's jogging speed (S), we will first find a common denominator and then solve the resulting equation: (3(S+40) + 5S) / (S(S+40)) = 0.5 3(S+40) + 5S = 0.5S(S+40) 3S + 120 + 5S = 0.5S^2 + 20S 0.5S^2 + 12S - 120 = 0 Now, we can either solve this quadratic equation by factoring or by using the quadratic formula. In this case, we will use the quadratic formula: S = (-b ± √(b^2 - 4ac)) / 2a S = (-12 ± √(144 + 240)) / 1 S = (-12 ± √384) / 1 S = 6 or S = -20 Since the speed cannot be negative, we take the positive value: Sam's jogging speed (S) = 6 km/h.

Key Concepts

Quadratic Equations

Quadratic equations are used to solve problems that involve unknowns squared in the equation (i.e., the equation has a term like \(x^2\)). These types of equations can commonly be found in various problem-solving scenarios, such as finding roots in mathematics or analyzing physics problems.
In our exercise, the quadratic equation arises when we set up the equation for Sam's jogging speed. We end up with an equation that looks like:

• \(0.5 S^2 + 12 S - 120 = 0\)

This equation is in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. Solving such equations typically involves factoring, completing the square, or using the quadratic formula:

• \(S = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}\)

This formula allows us to find solutions for the variable \(S\) in our equation. It's essential to remember that in practical scenarios like speed, negative results do not make sense, so we often disregard them. Solving quadratic equations requires careful calculation to avoid mistakes, particularly when dealing with square roots and signs.

Distance, Speed, and Time Problems

Distance, speed, and time problems involve understanding the relationship between these three core concepts. The key formula is:

• \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \)

This formula helps us relate how fast something goes with how far it travels and how long it takes. It's crucial to convert time into consistent units (such as hours) when calculating.
For instance, in the exercise, Sam jogs 3 kilometers, and then rides the train for 5 kilometers. We denote Sam's jogging speed as \(S\) and the time to jog as \(J\). Hence, for jogging, the formula becomes \(S = \frac{3}{J}\). Similarly, we apply the formula to the train ride. Understanding how to set these up allows us to solve complex problems efficiently. The key steps include:

• Identifying what you know and need to find out.
• Setting up the correct equations using the formulas.
• Solving these equations systematically.

Practicing these problems enhances your ability to work through real-world situations where speed and distances play crucial roles.

Problem Solving

Effective problem solving involves breaking down complex problems into manageable steps, identifying known and unknown variables, and systematically finding a solution. Our exercise showcases this nicely by focusing on a structured approach:

• First, understand the problem by identifying given information and what is required.
• Translate the word problem into mathematical equations using known formulas.
• Use logical steps to manipulate and solve these equations.

In Sam's scenario, we had to identify different speeds (jogging, train), distances, and times, and then represent these using variables. Following this, equations were setup that linked jogging speed to the train speed, and the whole into a time equation.
Problem solving in maths teaches us to keep track of variables, substitute where necessary, and apply techniques like elimination or factorization. Regular practice with different types of problems will improve your ability to tackle diverse and challenging scenarios easily.