Randy is walking home from school. According to the diagram above, what is his total distance from school to home? Show your work and include units. If he had a jet pack, would you use distance or displacement? Why?

if he needs to walk, we can see that between the street and his house he must walk 4 times a distance of 0.5km, so this is a total of 4¨*0.5km = 2km.

Now he has a jet-pack, he can ignore the buildings and just travel in the shorter path, so we can draw a triangle rectangle, in such a way that the hypotenuse of this triangle is the distance between the home and the school.

One of the cathetus is the vertical distance, in this case, is 1km, and the other one is the horizontal distance, also 1km.

So the actual distance is given by the Pythagorean's theorem:

A^2 + B^2 = H^2

Where A and B are the cathetus, and H is the hypotenuse, then:

H^2 = 1km^2 + 1km^2

H = (√2)km = 1.41km.

Now, in the case that he has a jet-pack, he can actually go to the school using this hypotenuse line as his path, so in this case the distance and the displacement would be the same.

Distance: "how much ground an object has covered"

Displacement: "Difference between the final position and the initial position"

When he walks, the distance is 2km, but the displacement is 1.41km

When he uses the jet-pack, both the distance and the displacement are 1.41km

the least common denominatore of 7 and 2 is 9

step-by-step explanation:

7+2=9