Problem 50 - Algebra A Combined Function

Question

On a 255 -mile trip, Gary Alessandrini traveled at an average speed of \(70 \mathrm{mph},\) got a speeding ticket, and then traveled at \(60 \mathrm{mph}\) for the remainder of the trip. If the entire trip took 4.5 hours and the speeding ticket stop took 30 minutes, how long did Gary speed before getting stopped?

Step-by-Step Solution

Answer

Gary sped for 1.5 hours before getting stopped.

Step 1: Express Total Driving Time

The entire trip took 4.5 hours. During this time, Gary spent 0.5 hours (30 minutes) stopped for a speeding ticket, so the actual driving time is 4.5 hours - 0.5 hours = 4 hours.

Step 2: Define Variables for Distance

Let \( d_1 \) be the distance Gary traveled at 70 mph, and let \( d_2 \) be the distance traveled at 60 mph. We know the total distance \( d_1 + d_2 = 255 \) miles.

Step 3: Express Distances in Terms of Time

The time for each part of the trip is given by the formula \( \text{time} = \frac{\text{distance}}{\text{speed}} \). Thus, \( d_1 = 70t_1 \) and \( d_2 = 60t_2 \) where \( t_1 \) and \( t_2 \) are the times spent at 70 mph and 60 mph respectively. Also, \( t_1 + t_2 = 4 \) hours.

Step 4: Solve Simultaneous Equations

We have two equations: 1. \( 70t_1 + 60t_2 = 255 \) (from distance)2. \( t_1 + t_2 = 4 \) (from time)Substitute \( t_2 = 4 - t_1 \) into the first equation: \[ 70t_1 + 60(4 - t_1) = 255 \]\[ 70t_1 + 240 - 60t_1 = 255 \]\[ 10t_1 = 15 \]\[ t_1 = 1.5 \] hours.

Step 5: Calculate Speeding Duration

Since \( t_1 = 1.5 \) hours, Gary sped for 1.5 hours before being stopped.

Key Concepts

Distance

Distance refers to how far an object travels. In our example, the entire journey Gary completed was a total of 255 miles.
Distance can be broken into segments, such as different speeds or conditions during the trip. Mathematically, distance is represented by the formula:

\[ \text{Distance} = \text{Speed} \times \text{Time} \]

This formula helps in calculating how far Gary traveled at 70 mph and 60 mph separately. When dealing with word problems involving distance, always look to split the total journey based on the conditions given, like changes in speed.
Finally, it's important to remember that in our case, the sum of different parts of Gary's trip (\(d_1\) and \(d_2\)) equals the total trip distance, which is 255 miles.

Speed

Speed is the rate at which someone or something moves. It plays a crucial role in determining how fast a distance is covered.
Speed in our case changes during Gary's trip. First, he drives at 70 mph, then at 60 mph after the speeding ticket. These changes impact how long it takes him to complete each section of the trip.

• At 70 mph, this represents a greater speed, meaning Gary covers more distance in less time.
• At 60 mph, the speed is less, meaning it takes more time to cover the same distance compared to 70 mph.

Using the concept of speed, remember that for a given distance, an increase in speed will decrease travel time and vice versa. This is an essential principle when solving problems involving distance and time.

Time

Time in this problem is all about how long each part of the journey takes. Gary’s total driving time was 4 hours, accounting for his 30-minute stop. This means he drove for 1.5 hours at 70 mph and the remainder at 60 mph.
Time for each segment of his trip is expressed using the formula:\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]
This allows you to calculate how much time was spent at each speed. The time for the combined segments of his trip was used to find how long he was speeding before getting stopped.

In the solution, solving the equations

• \( t_1 + t_2 = 4 \)
• \( 70t_1 + 60t_2 = 255 \)

helped to find the time at each speed, which were crucial in answering the exercise question.