Question

You are making a 350 -mile car trip. You decide to drive a little faster to save time. Choose an expression for the time saved if the car's average speed \(s\) is increased by 5 miles per hour. $$\begin{array}{lllll} \text { (A) } \frac{350}{s+5} & \text { (B) } \frac{s+5}{350}-\frac{s}{350} & \text { (C) } \frac{350}{s}-\frac{350}{s+5} & \text { (D } 350(s+5)-350 s \end{array}$$

Short answer

The correct expression for the time saved is (C) \( \frac{350}{s} - \frac{350}{s+5} \)

Step-by-step solution

Calculate trip time with initial speed

The length of the trip is 350 miles, and the car's initial average speed is \(s\) mph. The time it takes to complete the trip at the initial speed is given by \( time_{initial} = \frac{distance}{speed} = \frac{350}{s} \)

Calculate trip time with increased speed

When the speed is increased by 5 mph, the new average speed becomes \(s+5\) mph. The time it takes to complete the trip at the increased speed is \( time_{increased} = \frac{distance}{speed} = \frac{350}{s+5} \)

Calculate time saved

The time saved can be calculated by subtracting the time taken with the increased speed from the time taken with the initial speed. Therefore, the time saved is \( time_{saved} = time_{initial} - time_{increased} = \frac{350}{s} - \frac{350}{s+5} \)

Key concepts

Average Speed

Understanding average speed is crucial when calculating time saved during a trip. In simple terms, average speed is the total distance traveled divided by the total time taken to cover that distance. When you are driving, your speedometer tells you your instantaneous speed, but your average speed may be different because it accounts for all variations in speed during the trip, including stops.

For example, if you're making a trip that covers a total of 350 miles and you drive consistently at 50 miles per hour (mph), your average speed is also 50 mph. However, if you sometimes drive faster and sometimes get stuck in traffic, you might find that your average speed over the 350 miles is less than 50 mph. Algebraic expressions can be used to denote average speed in mathematical terms, like the expression 's' used in the textbook problem.

Distance Rate Time

The fundamental formula to remember when considering motion problems is the distance, rate, and time formula: \[ distance = rate \times time \]. This equation is derived from the basic principle that the distance covered is directly proportional to the speed and the amount of time traveled at that speed.

As it relates to our example, we have a fixed distance (350 miles) and we're manipulating the rate (or speed) by increasing it. It's essential to note that if the speed increases, the time taken for the trip will decrease. The direct relationship between rate and time in determining distance allows us to calculate the time saved by plugging into the formula different rates, namely the initial speed and the increased speed.

Algebraic Expressions

Algebraic expressions are mathematical phrases involving numbers, variables, and operation symbols. In the context of our textbook problem, symbols such as 's' are used to represent quantities that can change — in this case, the car's average speed. An increase in speed is represented as \(s+5\), where 5 represents the additional mph.

The beauty of algebra is in its ability to solve problems by using unknown variables. The expression for time saved, \(\frac{350}{s} - \frac{350}{s+5}\), utilizes algebraic manipulation to illustrate the difference in travel time resulting from a speed change. This shows the shift in time due to the increased speed and allows students to visualize the concept through the lens of algebra.