What was the net distance traveled by my dog?

Question

My dog ran out of the house and it took . one hour an a half later I caught the dog, at a park three miles East and four miles South of my house. What was the net distance traveled by my dog? What was my dog's average speed?

Answer 1

You can define a right triangle with the sides 3 (east) and 4 (south) and solve for the hypotenuse. A^2 + B^2 = C^2. 3, 4, 5 is a well known right-triangle, actually. So the net distance is 5 miles, southeast (I can solve for the exact angle, too, but that will take a little longer. Let me know if you need it.) Your dog's average speed? Well, that depends on what route he took, and how far he actually travelled. If he went 3 miles east, and then 4 miles south, he travelled 7 miles. But to give you an answer, I'll assume he travelled in a straight line, 5 miles. So 5 divided by 3/2 hours, same as 5 times 2/3, is 10/3 or about 3.33 miles per hour.

Answer 2

The magnitude of the net displacement (5 miles) is easily calculated by using Pythagorus.

The magnitude of the average velocity is easily calculated by dividing that by an hour and a half. (3.333 miles per hour).

The average speed, however, will be greater than that. No dog I've ever met travels anywhere in a straight line. They bop here and there and everywhere to pee on everything. These excursions cancel out if you are calculating average velocity (a vector quantity), but they matter if you are calculating average speed.

If you can explain that to your teacher, I promise you an A++ for the question.

Answer 3

i was going to ask you that question..lol
9m/hr for speed
12.6 m