Problem 306 - Elementary Algebra

Question

Sarah wants to arrive at her friend's wedding at 3:00. The distance from Sarah's house to the wedding is 95 miles. Based on usual traffic patterns, Sarah predicts she can drive the first 15 miles at 60 miles per hour, the next 10 miles at 30 miles per hour, and the remainder of the drive at 70 miles per hour. (a) How long will it take Sarah to drive the first 15 miles? (b) How long will it take Sarah to drive the next 10 miles? (c) How long will it take Sarah to drive the rest of the trip? (d) What time should Sarah leave her house?

Step-by-Step Solution

Answer

Sarah should leave at 1:25 PM.

Step 1: Calculate Time for First 15 Miles

Use the formula \text{time} = \frac{\text{distance}}{\text{speed}}. For the first 15 miles at 60 miles per hour:\( t_1 = \frac{15 \text{ miles}}{60 \text{ miles per hour}} = 0.25 \text{ hours} \text{ or } 15 \text{ minutes} \)

Step 2: Calculate Time for the Next 10 Miles

For the next 10 miles at 30 miles per hour, use the same formula:\( t_2 = \frac{10 \text{ miles}}{30 \text{ miles per hour}} = \frac{1}{3} \text{ hours} \text{ or } 20 \text{ minutes} \)

Step 3: Calculate Time for the Remaining Distance

First, determine the remaining distance after driving the first 25 miles (15 + 10):\( 95 \text{ miles} - 25 \text{ miles} = 70 \text{ miles} \)Then, calculate the time to drive these 70 miles at 70 miles per hour:\( t_3 = \frac{70 \text{ miles}}{70 \text{ miles per hour}} = 1 \text{ hour} \)

Step 4: Total Travel Time

Add the travel times from Steps 1-3:\( t_{\text{total}} = t_1 + t_2 + t_3 = 0.25 \text{ hours} + 0.33 \text{ hours} + 1 \text{ hour} = 1.58 \text{ hours} \text{ or } 1 \text{ hour and } 35 \text{ minutes} \)

Step 5: Calculate Departure Time

Sarah needs to arrive at 3:00 PM. Subtract the total travel time from the arrival time:\( 3:00 \text{ PM} - 1 \text{ hour and } 35 \text{ minutes} = 1:25 \text{ PM} \)

Key Concepts

distance

Distance is a measure of how far two points are from each other. In this problem, the total distance Sarah must travel to her friend's wedding is 95 miles. Distance is typically measured using units such as miles or kilometers. Understanding how to calculate distances covered in different segments of a journey, as Sarah does with 15 miles, 10 miles, and 70 miles, helps in accurately planning trips.

speed

Speed is the rate at which an object covers distance. It is usually expressed in units such as miles per hour (mph) or kilometers per hour (km/h). In Sarah's case, she predicts her speeds based on traffic patterns: 60 mph for the first 15 miles, 30 mph for the next 10 miles, and 70 mph for the remaining 70 miles. The formula used to find speed is: \(\text{speed} = \frac{\text{distance}}{\text{time}}\).

time calculation

Time calculation involves working out how long a journey or an activity will take. For Sarah's trip to the wedding, we use the formula: \(\text{time} = \frac{\text{distance}}{\text{speed}}\). For each segment of her trip, we calculate:

• For the first 15 miles at 60 mph: \( \frac{15 \text{ miles}}{60 \text{ mph}} = 0.25 \text{ hours} \text{ or } 15 \text{ minutes} \)
• For the next 10 miles at 30 mph: \( \frac{10 \text{ miles}}{30 \text{ mph}} = 0.33 \text{ hours} \text{ or } 20 \text{ minutes} \)
• For the last 70 miles at 70 mph: \( \frac{70 \text{ miles}}{70 \text{ mph}} = 1 \text{ hour} \)

Adding up all these times gives the total travel time.

departure time

The departure time is the time at which you need to start your journey to reach your destination on time. For Sarah, we need to calculate this based on her total travel time and when she needs to arrive. Since she needs to reach the wedding by 3:00 PM, and it takes her 1 hour and 35 minutes to travel, we subtract the travel time from the scheduled arrival time:
3:00 PM - 1 hour and 35 minutes = 1:25 PM.
This means Sarah should leave her house at 1:25 PM to arrive on time for her friend's wedding.