Question

A runs $\dfrac{3}{2}$ times as fast as B. If A gives B a start of 40 m, how far must the winning post be from the starting point, so that A and B reach at the same time?

Hint:

When one runner gives a head start, equate times using adjusted distances. Use $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$ and simplify algebraically.

Answer 1

Let B's speed be $v$ m/s. A's speed is $\dfrac{3}{2}v$ m/s. Let the distance to the winning post be $d$ meters.
A starts 40 m behind the starting line, so A runs $(d - 40)$ meters. B runs the full distance $d$ meters.
As A and B finish simultaneously, their times are equal:
\[\text{Time}_A = \text{Time}_B \Rightarrow \frac{d - 40}{\frac{3}{2}v} = \frac{d}{v}\]
Simplifying the equation:
\[\frac{2(d - 40)}{3v} = \frac{d}{v}\]
Multiply both sides by $3v$:
\[2(d - 40) = 3d\]
Expand and solve for $d$:
\[2d - 80 = 3d \Rightarrow 3d - 2d = -80 \Rightarrow d = -80\]
There seems to be an error in the initial setup. Let's re-evaluate the problem statement.
If A gives B a 40 m head start, this means B starts 40 m ahead of A. Let's assume A starts at the origin (0 m) and B starts at the 40 m mark.
Let A's starting point be 0. Then the winning post is at distance $d$.
B starts at 40 m, so B runs $(d - 40)$ meters to reach the winning post.
A runs the full distance $d$ meters.
Time taken by A: $\dfrac{d}{\frac{3}{2}v} = \dfrac{2d}{3v}$
Time taken by B: $\dfrac{d - 40}{v}$
Equating the times:
\[\dfrac{2d}{3v} = \dfrac{d - 40}{v}\]
Multiply both sides by $3v$:
\[2d = 3(d - 40)\]
Expand and solve for $d$:
\[2d = 3d - 120 \Rightarrow 3d - 2d = 120 \Rightarrow d = 120\]
The winning post is 120 m from A's starting point. B starts 40 m ahead, so B's starting point is at 40 m. The winning post is at 120 m.
Total distance from B's start to winning post = $120 - 40 = 80$ m. This is inconsistent with the prior formulation where B runs $d$. Let's clarify the starting positions.
Let the starting point for B be 0. The winning post is at $d$.
A gives a 40 m head start to B. This means A starts 40 m behind B's starting line. So A starts at $-40$ m and B starts at 0 m.
Distance A runs = $d - (-40) = d + 40$ meters.
Distance B runs = $d$ meters.
Time taken by A: $\dfrac{d + 40}{\frac{3}{2}v} = \dfrac{2(d + 40)}{3v}$
Time taken by B: $\dfrac{d}{v}$
Equating the times:
\[\dfrac{2(d + 40)}{3v} = \dfrac{d}{v}\]
Multiply both sides by $3v$:
\[2(d + 40) = 3d\]
Expand and solve for $d$:
\[2d + 80 = 3d \Rightarrow 3d - 2d = 80 \Rightarrow d = 80\]
The winning post is 80 m from B's starting point. The total distance of the race from A's starting point is $80 + 40 = 120$ m.
Therefore, the winning post is 120 m from A's start. (B starts at 40 m from A's start, and the winning post is at 120 m from A's start).