Problem 50 - Basic Technical Mathematics with Calculus

Question

Solve the given problems. All numbers are accurate to at least two significant digits. A student cycled \(3.0 \mathrm{km} / \mathrm{h}\) faster to college than when returning, which took 15 min longer. If the college is \(4.0 \mathrm{km}\) from home, what were the speeds to and from college?

Step-by-Step Solution

Answer

The speed to college was 8.6 km/h and returning was 5.6 km/h.

Step 1: Define the Variables

Let's define two variables for the speeds. Let \( v_1 \) represent the speed to college and \( v_2 \) represent the speed returning from college. According to the problem, the relation between the two speeds is \( v_1 = v_2 + 3.0 \).

Step 2: Set Up the Equation for Time

The time taken to cycle to college can be written as \( \frac{4.0}{v_1} \), and the time taken for returning as \( \frac{4.0}{v_2} \). Since it took 15 minutes longer to come back, we have: \( \frac{4.0}{v_2} = \frac{4.0}{v_1} + 0.25 \).

Step 3: Substitute the Speed Relationship

Substitute \( v_1 = v_2 + 3.0 \) into the time equation: \[ \frac{4.0}{v_2} = \frac{4.0}{v_2 + 3.0} + 0.25 \].

Step 4: Solve the Equation

Multiply every term by \( v_2(v_2 + 3.0) \) to eliminate the fractions: \[ 4.0(v_2 + 3.0) = 4.0v_2 + 0.25v_2(v_2 + 3.0) \]. Simplify and solve for \( v_2 \): \[ 12.0 = 0.25v_2^2 + 0.75v_2 \].

Step 5: Solve the Quadratic Equation

Rewrite the equation as \( 0.25v_2^2 + 0.75v_2 - 12.0 = 0 \). Multiply through by 4 to clear decimals: \( v_2^2 + 3v_2 - 48 = 0 \). Use the quadratic formula: \( v_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 3 \), \( c = -48 \).

Step 6: Calculate the Values

Calculate the discriminant: \( b^2 - 4ac = 3^2 - 4 \times 1 \times (-48) = 9 + 192 = 201 \). Thus, \( v_2 = \frac{-3 \pm \sqrt{201}}{2} \). Calculate the positive root as it's the realistic scenario for speed: \( v_2 \approx 5.6 \text{ km/h} \).

Step 7: Find the Speed to College

Using \( v_1 = v_2 + 3.0 \), we find \( v_1 = 5.6 + 3.0 = 8.6 \text{ km/h} \).

Key Concepts

Speed Calculations

Understanding how to calculate speed is crucial for solving problems related to movement over a distance. Speed is defined as the distance covered divided by the time taken to cover that distance. Mathematically, it can be expressed as:

• Speed = Distance / Time

Knowing how to manipulate this formula is key in solving various problems involving speed, distance, and time. In the provided problem, we needed to calculate two speeds: the speed to college (\(v_1\)) and the speed returning from college (\(v_2\)). We used these speed formulas along with additional given information to form equations that helped us solve for the unknowns.
Accuracy in setting up the problem is critical. For example, converting time appropriately (like changing 15 minutes to 0.25 hours) ensures that all units match, making the math consistent and allowing for correct solutions. Also, given the speeds involved in cycling to and from college, knowing the relationship between them (\(v_1 = v_2 + 3.0\)) was essential in setting up our equation for time.

Problem Solving

Problem-solving in math involves several systematic steps that, when followed, can simplify even the most complex problems. In this exercise, the first step was to define variables that represent unknowns or elements of the problem. Here, identifying speeds as unknown variables (\(v_1\) and \(v_2\)) allowed us to frame our problem in mathematical terms.
After defining the variables, setting up equations that incorporate the relationships given in the problem statement comes next. For instance, since the return trip took longer, this difference in time was transcribed into an equation involving both speeds:

• \(\frac{4.0}{v_2} = \frac{4.0}{v_1} + 0.25\)

This approach—systematically translating words into math—transforms what might seem like a wordy, convoluted problem into a clear, solvable equation.
Solving the resulting equations requires knowledge of algebra and sometimes calculus, but with practice, these steps become intuitive. For this problem, solving a quadratic equation was necessary to find realistic speed values, emphasizing the importance of mathematical skills in problem-solving.

Distance and Time Relationship

The connection between distance, speed, and time is pivotal for understanding motion in physics and everyday problem-solving. By grasping how these three variables interact, one can make informed predictions and solutions.
At its core, the idea is straightforward:

• When speed increases, and the distance stays the same, the time required decreases.
• Conversely, if the speed decreases, more time is needed to cover the same distance.

In our problem, this relationship explained why the trip to college was faster while returning took longer; it was because the cycling speeds differed, affecting the time spent on the road.
Understanding these interactions allows problem solvers to set up equations that model real-world situations effectively. For example, altering the speed affected the return time by a known amount, represented by reducing the equation to its essence, \(\frac{4.0}{v_2} = \frac{4.0}{v_1} + 0.25\), which in turn could be solved for specific values of speed. Working through such equations highlights how central this relationship is across many mathematical contexts.