Doug is a dog, and his friend Bert is a bird. They live in Salt Lake City, where the streets are 1/16 miles apart and arranged in a

square grid (see map). They are both standing at the corner of 6th avenue and L street. Doug can run at an average speed of 30

mi/hr through the streets of Salt Lake, and Bert can fly at an average speed of 20 mi/hr. They are about to race to the corner of

10th avenue and Estreet. who would win and why?

Doug wins. He finishes in 1.38 minutes. Bert takes 1.51 minutes.

Step-by-step explanation:

Each city block is a square 1/16 mile by 1/16 mile.

From 6th Av and L Street to 10th Av and E Street, they need to go 4 avenues and 7 streets over. These two distances along blocks are the legs of a right triangle. The right triangle has legs of length 4 * 1/16 mile and 7 * 1/16 mile.

leg1 = 4 * 1/16 mile = 0.25 mile

leg2 = 7 * 1.16 mile = 0.4375 mile

The hypotenuse of the right triangle can be calculated by the Pythagorean theorem.

a^2 + b^2 = c^2

(0.25 mile) ^2 + (0.4375 mile) ^2 = c^2

c = 0.50389

The direct distance from one corner to the other is 0.50389 miles.

The bird travels this distance since the bird can fly above the streets.

The dog has to travel along the legs of right triangle by running in the streets and avenues. The dog travels 0.25 mile + 0.4375 mile = 0.6875 miles

Now we need to see what takes less time:

Bird: distance of 0.50389 miles at 20 mph, or

Dog: distance of 0.6875 miles at 30 mph

We start with the formula for speed:

speed = distance/time

s = dt

Now we solve it for time:

st = d

t = d/s

Time equals distance divided by speed.

Now we find the times for the bird and for the dog.

Bird:

t = d/s = (0.50389 mi) / (20 mi/h) = 0.02519 hours = 1.51 minutes

Dog:

t = d/s = (0.6875 mi) / (30 mi/h) = 0.0229 hours = 1.38 minutes

The bird takes 1.51 minutes, and the dog takes 1.38 minutes.

The dog takes less time to get there, so the dog wins.

No map given but it shouldn't matter.

E is the 5th letter, L the 12th.

Start at S (6,12). End at E (10,5)

The vector between them, E-S = (4,-7)

Each unit is 1/16 of a mile, though that probably doesn't matter that much.

Doug has to stay on the grid, so has to run |4|+|7|=11 units. At 30 mi/hr that takes (11/16) / 30 = 0.022916 hours.

Bert can go diagonally, so flies √ (4²+7²) = √65 ≈ 8.06 units. At 20 mi/hr that takes (8.06/16) / 20 = 0.025194 hours.

Answer: Doug wins

Why? Because it's quicker to cover 4+7 at 30 mph than it is to cover √ (4²+7²) at 20 mph. That is, Doug is 1.5 times faster and the 1.5 times the diagonal distance is more than the grid distance.