Problem 61 - Calculus with Concepts in Calculus

Question

Starting at noon, \(A\) flies 2400 miles from New York to San Francisco at a velocity of 400 miles per hour. \(B\) starts the same trip at \(2: 00\) P.M. the same day with a velocity of 800 miles per hour. Express the distance \(D\) between \(A\) and \(B\) at any instant between noon and \(5: 00 \mathrm{P} . \mathrm{M}\). in terms of the time in hours elapsed after noon.

Step-by-Step Solution

Answer

The distance between A and B, in terms of hours elapsed after noon, is \(|400t - 1600|\).

Step 1: Determine Expression for A's Distance

Since A starts traveling at noon, the distance traveled by A in time \( t \) (where \( t \) is the time in hours since noon) is described by the formula \( D_A(t) = 400t \) miles.

Step 2: Determine Expression for B's Distance

Because B starts traveling at 2:00 PM, B has only traveled for \( t - 2 \) hours at any time \( t \) (where \( t \) is more than 2). The distance B has traveled is \( D_B(t) = 800(t - 2) \) miles.

Step 3: Expression for Distance Between A and B

The distance between A and B is the absolute value of the difference between their distances from New York. Therefore, \( D(t) = |D_A(t) - D_B(t)| = |400t - 800(t - 2)| \).

Step 4: Simplify the Distance Expression

First, expand the expression: \( 800(t - 2) = 800t - 1600 \). Now, \( D(t) = |400t - (800t - 1600)| = |400t - 800t + 1600| = |-400t + 1600| = |400t - 1600| \).

Step 5: Final Expression of Distance Between A and B

The distance between A and B in terms of time since noon \( t \) is \( D(t) = |400t - 1600| \).

Key Concepts

Distance Calculation

Distance calculation is understanding how far an object travels over a period of time. It involves the use of basic math to find this distance based on speed and time. In the exercise, the first step is to determine the distance traveled by each plane, A and B. We denote distance as \( D_A(t) \) for plane A and \( D_B(t) \) for plane B. Here’s how you calculate it:

• For A traveling at 400 miles per hour, the distance can be found as \( D_A(t) = 400t \).
• For B, which starts 2 hours later, the formula becomes \( D_B(t) = 800(t - 2) \), as B only travels after he begins its journey.

Distance calculation is a foundational step in solving problems involving motion. Once you get the hang of it, tackling other areas like velocity and algebra becomes easier. It's about knowing the speed and start time, then applying the formula to find distance.

Relative Velocity

Relative velocity is key when two objects are moving towards, away from, or in parallel to each other. It helps us understand how the two speeds relate to each other. Consider it as a way to measure how fast one is approaching the other. In this exercise, we’re interested in how the distance between A and B changes over time:
For A and B:

• A moves at a constant speed of 400 miles per hour.
• B starts later but at a faster speed of 800 miles per hour.

Relative velocity helps determine how quickly the gap between them widens or narrows. By calculating the difference in distances \(|D_A(t) - D_B(t)|\), you apply the concept of relative velocity to understand their positions over time. Here, we realize that even with different start times and speeds, their motion can be compared effectively using the concepts of relativity in motion.

Algebraic Expressions

Algebraic expressions are crucial when solving calculus word problems. They simplify and help us model the real-world movement mathematically to find solutions. In our problem, expressions model the movement of A and B over time and find the distance between them:

• The expression for A, \( 400t \), represents the distance A covers.
• B’s expression is \( 800(t - 2) \), recognizing its delayed start time.
• To find the distance between A and B, use \( |400t - 800(t - 2)| \).

Simplifying algebraic expressions like these involves distributing terms, combining like elements, and sometimes using absolute value notation to handle differing speeds and start times. Such expressions allow for calculating precise distances over variable time frames without guesswork, only logic and math. That precision is the power of algebra in motion problems.