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 the speed of B be $v$ m/s.
Then the speed of A = $\dfrac{3}{2}v$ m/s (as A runs $\dfrac{3}{2}$ times faster).
Let the total distance from starting point to winning post be $d$ meters.
B runs the full distance $d$, while A runs only $(d - 40)$ meters, since A gives a 40 m head start.
Since both A and B reach the finish line at the same time, their times are equal:
\[ \text{Time}_A = \text{Time}_B \Rightarrow \frac{d - 40}{\frac{3}{2}v} = \frac{d}{v} \]
Multiply both sides by $\frac{3}{2}v$ to eliminate the denominator on the left:
\[ d - 40 = \frac{3}{2}d \]
Now multiply both sides by 2:
\[ 2(d - 40) = 3d \Rightarrow 2d - 80 = 3d \]
Bring all terms to one side:
\[ 2d - 80 - 3d = 0 \Rightarrow -d = 80 \Rightarrow d = 80 \]
Wait — this seems off. Let’s redo the algebra correctly.
From previous step: \[ d - 40 = \frac{3}{2}d \Rightarrow \text{Multiply both sides by 2: } 2(d - 40) = 3d \Rightarrow 2d - 80 = 3d \Rightarrow 3d - 2d = 80 \Rightarrow d = 80 \]
Wait again — our initial logic was reversed. If A gives a head start, B covers $(d)$, A covers $(d - 40)$.
Let’s revise.
Time taken by A: $\dfrac{d - 40}{\frac{3}{2}v} = \dfrac{2(d - 40)}{3v}$
Time taken by B: $\dfrac{d}{v}$
Now equate: \[ \dfrac{2(d - 40)}{3v} = \dfrac{d}{v} \Rightarrow 2(d - 40) = 3d \Rightarrow 2d - 80 = 3d \Rightarrow d = 80 \]
So, winning post must be 120 m from A's start (B's start = 0, A starts at 40 m mark).
But if B starts 40 m ahead, A runs $(d - 40)$ and B runs $d$.
Total distance from B's start to winning post = $d = 120$