Question

Exercises \(57-59\), you are going on a trip to the Natural History Museum. At 9:00 A.M., you leave for the museum, which is 120 miles away. At 10: 15 A.M., you are 63 miles away from the museum. According to your equation, when will you reach the museum?

Short answer

According to calculations, arrival at the museum is expected at around 11:38 A.M.

Step-by-step solution

Understand the problem

The problem provides the total distance to the museum (120 miles), the distance left to reach the museum at a certain time (63 miles at 10:15 A.M.) and the start time of the journey (9:00 A.M.). We need to find out the time of arrival at the museum.

Calculate the average speed

Firstly, calculate the speed for the first part of the trip. The distance covered until 10:15 A.M. is \(120 - 63 = 57\) miles. The time taken to cover this distance was from 9:00 A.M. to 10:15 A.M., which is \(1 \frac{1}{4}\) hours or 1.25 hours. So, the average speed is \(\frac{distance}{time} = \frac{57}{1.25} = 45.6\) miles per hour.

Calculate the time to reach the museum

To find out when the arrival at the museum will be, calculate the time it would take to cover the remaining distance at this average speed. The remaining distance is 63 miles, and if we travel this distance at a speed of 45.6 miles per hour, the time would be \(\frac{distance}{speed} = \frac{63}{45.6} = 1.38\) hours. Since this is added to the time 10:15 A.M., the arrival time would be at nearly 11:38 A.M.

Key concepts

Average Speed Calculation

Understanding how to calculate average speed is crucial in solving many algebraic word problems that involve movement, and it's especially important when planning trips. The average speed is simply the total distance traveled divided by the total time taken to travel that distance. In mathematical terms, the formula for average speed (\( v_{avg} \) is: \[ v_{avg} = \frac{total\ distance}{total\ time} \].

For instance, when you're traveling to a destination and you've covered a certain distance over a specific time period, you can determine your average speed by using this formula. Remember, when you're working out the time, it's best to convert it into hours if your distance is in miles, or into seconds if your distance is in meters, to maintain a consistent unit of measure. This becomes very handy when measuring how fast you'll need to go to reach a destination at a planned time.

Time-Distance Relationship

The time-distance relationship is an essential concept in both mathematics and everyday life, helping us to understand how time and distance correlate with speed. The core idea is simple: the distance covered is directly proportional to the speed and the time taken, which can be represented by the formula \[ distance = speed \times time \].

In the context of traveling to a place like the Natural History Museum, calculating the remaining distance you need to travel, and knowing your speed, allows you to estimate the time it will take to get there. Conversely, if you know how much time you have and your speed, you can determine how far you'll travel. Manipulating this relationship gives us powerful insights into planning and estimation for travel and logistics. Using unit conversion and clear understanding of this relationship, one can solve many real-life problems with ease.

Problem-Solving Strategies

When facing algebraic word problems, having a good problem-solving strategy is key. Breaking down the problem into manageable steps can turn a daunting task into a series of simpler calculations.

The first step, as shown in the exercise, is always to understand the problem. Ask yourself what is being asked and identify the information given. Next, translate the problem into mathematical expressions or equations. This might mean calculating averages, like speed, or applying the time-distance relationship. Once you have your equations, solve them step by step, and finally, interpret the solution in the context of the problem to ensure it makes sense. Logical reasoning and organized work can make complex problems far more approachable.