Problem 19 - Algebra and Trigonometry with Analytic Geometry

Question

At 6 A.M. a snowplow, traveling at a constant speed, begins to clear a highway leading out of town. At \(8 \mathrm{~A} . \mathrm{M}\). an automobile begins traveling the highway at a speed of \(30 \mathrm{mi} / \mathrm{hr}\) and reaches the plow \(30 \mathrm{minutes}\) later. Find the speed of the snowplow.

Step-by-Step Solution

Answer

The speed of the snowplow is 6 miles per hour.

Step 1: Define Variables

Let the speed of the snowplow be \( v \) miles per hour. We know the car travels at \( 30 \) miles per hour.

Step 2: Calculate Time the Snowplow was on the Highway

The snowplow began at 6 AM and the car caught it at 8:30 AM. So, the snowplow has been traveling for a total of \( 8.5 - 6 = 2.5 \) hours.

Step 3: Calculate Distance Traveled by the Car

The car started at 8 AM and reached the snowplow at 8:30 AM, traveling for 0.5 hours. Its distance is \( 30 \times 0.5 = 15 \) miles.

Step 4: Set the Equations Equal

Since both the snowplow and car traveled the same distance by the time they meet, set the distances equal: \( 2.5v = 15 \).

Step 5: Solve for the Speed of the Snowplow

Solve the equation from Step 4 to find \( v \): \( 2.5v = 15 \) implies \( v = \frac{15}{2.5} = 6 \).

Step 6: Interpret the Result

The speed of the snowplow is \( 6 \) miles per hour.

Key Concepts

Speed Calculation

Speed plays a crucial role in understanding how quickly an object is moving over time. In our example, the snowplow's speed is what we aim to find, given certain conditions. The basic formula for speed is:

• Speed = Distance / Time

This formula helps us determine how fast the snowplow is moving by dividing the total distance it covers by the time it takes to cover that distance.
To calculate the snowplow's speed, we first need to determine both the time and the distance involved. In our exercise, we gathered that the snowplow began at 6 A.M. and was surpassed by the car at 8:30 A.M., indicating the snowplow was in motion for 2.5 hours. The car traveled 15 miles in 0.5 hours to reach the snowplow. With these elements, we can set up an equation to solve for the snowplow's speed.

Distance-Time Relationship

The relationship between distance and time is key to solving many algebra word problems. In essence, this relationship describes how far an object travels over a certain period at a given speed. The fundamental equation encapsulating this is:

• Distance = Speed × Time

Using this framework, we can analyze movements like the snowplow's. It allows us to calculate distances when speeds and time intervals are known, or inversely, to deduce speeds because of distance and time, as we did in our scenario.
When the car travels and catches up with the snowplow, both have covered the same distance when they meet. Therefore, we set the distances equal: the car's distance in 0.5 hours equals the snowplow's distance in 2.5 hours. This helps us formulate an equation to find an unknown speed, illustrating the balance between distance and time.

Linear Equations

Linear equations are fundamental to algebra, representing constant rates. In these equations, we solve for a variable by balancing both sides. The exercises often involve finding unknowns through a step-by-step solution.
In our snowplow problem, identifying the speed involves setting up a linear equation. We equate the distance traveled by the snowplow with that of the car, since both were equal at the point they met:

• 2.5v = 15

In this equation:

• 2.5 represents the time the snowplow was traveling.
• v is the snowplow's speed.
• 15 is the distance both traveled by the meeting time.

To solve for the speed, or the unknown variable, we divide both sides by 2.5, yielding a speed of 6 miles per hour.
Linear equations like this are powerful tools for revealing the dynamics and relationships in algebra word problems, often reducing complex scenarios to simple calculations.