Question

A runs \(\frac{7}{4}\) times as fast as \(B\). If \(A\) gives \(B\) a start of \(300 \mathrm{~m}\), how far must the winning post be if both \(A\) and \(B\) have to end the race at same time? (a) \(1400 \mathrm{~m}\) (b) \(700 \mathrm{~m}\) (c) \(350 \mathrm{~m}\) (d) \(210 \mathrm{~m}\)

Short answer

Answer: 700m

Step-by-step solution

Assign Variables

Assign variables to the known quantities. Let the speed of B be denoted as \(v\), and the speed of A as \(\frac{7}{4}v\). Let the total distance of the race be denoted as \(d\) and the head start given to B as \(300m\). We're looking for \(d\).

Understand the Relationship between Speed, Distance, and Time

The formula that connects speed, distance and time is \(speed = \frac{distance}{time}\), or we can rewrite it as \(time = \frac{distance}{speed}\). Since A and B finish the race at the same time, they both travel for the same amount of time. Therefore, we have the equation: \(\frac{d}{\frac{7}{4}v} = \frac{d-300}{v}\)

Solve the equation

First multiply both sides of the equation by \(v\) to eliminate the fraction on the right-hand side: \({\frac{4}{7}}d = d - 300\).

Rearrange the equation

Rearrange the equation to solve for \(d\) by moving \({\frac{4}{7}}d\) from the left side to the right side: \(d - {\frac{4}{7}}d = 300\). This simplifies to \({\frac{3}{7}}d = 300\).

Evaluate for \(d\)

To solve for \(d\), multiply both sides of the equation by \(\frac{7}{3}\): \(d = \frac{7}{3} \times 300 = 700\). Therefore, the winning post must be \(700m\) away. So, option (b) \(700m\) is the correct answer.

Key concepts

Speed of A and B

In this problem, we are dealing with two runners, A and B, with different speeds. Runner A is faster than runner B. Specifically, A runs

• \(\frac{7}{4}\) times the speed of runner B.

To break this down, if B’s speed is denoted as \(v\), then A’s speed would be \(\frac{7}{4}v\). Understanding their speeds is crucial because it dictates how much ground each runner can cover in a given time period.
Understanding the speeds in comparative terms helps us identify how they will perform relative to each other over the same duration. This forms the foundation for setting up the equations necessary to solve problems related to races and distances.

Equation Solving

When faced with problems where two variables need to cover different distances in the same time, setting up an equation is essential. Let’s dissect how that works here.
We start by using the formula for time:

• \(time = \frac{distance}{speed}\)

Since A and B finish the race at the same time, we equate their times of travel:

• \(\frac{d}{\frac{7}{4}v} = \frac{d-300}{v}\)

Our goal is to solve for \(d\), the total race distance. After isolating the variables and simplifying the equation, by multiplying and rearranging, we find:

• \(d - \frac{4}{7}d = 300\)
• \(\frac{3}{7}d = 300\)

Bringing everything together, we use simple arithmetic to find \(d\) by multiplying both sides by \(\frac{7}{3}\), yielding \(d = 700\). Equation solving allows us to transform a complex problem into manageable math.

Races and Head Starts

Head starts in races are an interesting twist because they allow a slower competitor to catch up. In this scenario, B starts 300 meters ahead of A. This head start is a crucial factor because it compensates for B's slower speed.
The head start effectively shifts the starting line for B, giving them an advantage they need to maintain the possibility of finishing at the same time as A. In practice, calculating the effect of a head start involves reducing the total distance B covers by the head start amount. Thus, B only needs to cover \(d - 300\) meters to finish the race.
Understanding head starts allows us to adjust the expectations and predictions of how such races will play out, considering each competitor’s unique advantages.