Two people set out from A along the straight road AB at speeds of 1.5 m/s and 2.0 m/s. The second person reaches B 5.5 minutes before the first. How long is AB?

Step-by-step solution

1. 1

   Name the unknown and write both times. Let $s$ be the length of $AB$ in metres. Both walkers cover the same distance, so $t_1=\frac{s}{1.5},\qquad t_2=\frac{s}{2}$

2. 2

   Convert the time gap to seconds. The speeds are in $\text{m/s}$, so the gap must be in seconds too: $5.5\text{ min}=5.5\times 60=330\text{ s}$

3. 3

   Set up the equation from "arrives earlier". The slower walker takes longer, so the difference $t_1-t_2$ is the head start: $\frac{s}{1.5}-\frac{s}{2}=330$

4. 4

   Combine the fractions. $\dfrac{s}{1.5}=\dfrac{2s}{3}$, so $\frac{2s}{3}-\frac{s}{2}=\frac{4s-3s}{6}=\frac{s}{6}=330$

5. 5

   Solve and verify. $s=6\times 330=1980\text{ m}$. Checking: $1980/1.5=1320\text{ s}$ and $1980/2=990\text{ s}$, and $1320-990=330\text{ s}=5.5\text{ min}$, exactly as required. The answer is B.