Problem 413

Solution 1

Suppose that when Nate arrives on time, his drive takes $t$ hours.

When Nate arrives 1 hour early, he arrives in $t-1$ hours.

When Nate arrives 1 hour late, he arrives in $t+1$ hours.

Since the distance that he drives is the same in either case and distance equals speed multiplied by time, then $\text{(60 km/h ) ((t-1) h ) = (40 km/h ) ((t+1) h )}$.

Expanding, we obtain $60t - 60 = 40t + 40$ and so $20t = 100$ or $t = 5$.

The total distance that Nate drives is thus $\text{(60 km/h ) (4 h ) = 240 km}$.

When Nate drives this distance in 5 hours at a constant speed, he should drive at $\text{240 km 5 h}$ which equals $\text{48 km/h}$.

Solution 2

Suppose that the distance that Nate drives is $d$ km.

Since driving at 40 km/h causes Nate to arrive 1 hour late and driving at 60 km/h causes Nate to arrive 1 hour early, then the difference between the lengths of time at these two speeds is 2 hours.

Since time equals distance divided by speed, then $$\text{d km 40 km/h - d km 60 km/h = 2 h}$$ Multiplying both sides of the equation by 120 km/h, we obtain $$\text{3d km - 2d km = 240 km}$$ which gives us $d = 240$.

Thus, the distance that Nate drives is 240 km.

At 40 km/h, the trip takes 6 hours and Nate arrives 1 hour late.

To arrive just in time, it should take 5 hours.

To drive 240 km in 5 hours, Nate should drive at a constant speed of $\text{240 km 5 h = 48 km/h}$.