Travis walked 20 yards north from the swing set to the slid. He then went 30 yards east to get a drink of water. From there he ran 40 yards south to play basketball. Finally, he ran 30 yards west to his bike. How far was hisbike from the swingset? I came up with 120 yards is that right?
He went 30 yards East and 30 yards West, so they cancel each other out. Collapse the diagram and just consider the N/S travel.
He started out from the swingset to go 20 yards North, and then went 40 yards South, which was parallel to where the bike was.
Do you think you can work it from here?
I hope this helps. Thanks for asking.
To find the distance between Travis's bike and the swing set, we need to calculate the total displacement from the swing set to the bike.
Travis walked 20 yards north from the swing set to the slide, so we can consider this as the positive y-direction.
Next, he went 30 yards east to get a drink of water, so we can consider this as the positive x-direction.
From there, he ran 40 yards south to play basketball, resulting in a negative y-direction displacement.
Finally, he ran 30 yards west to his bike, which corresponds to a negative x-direction displacement.
To find the total displacement, we can calculate the net change in both the x and y-directions.
Net change in the x-direction: 30 yards (east) - 30 yards (west) = 0 yards
Net change in the y-direction: 20 yards (north) - 40 yards (south) = -20 yards
Now, to find the distance between the swing set and the bike, we can use the Pythagorean theorem. The distance formula is √(Δx² + Δy²), where Δx is the net change in the x-direction and Δy is the net change in the y-direction.
Distance = √(0² + (-20)²) = √(0 + 400) = √400 = 20 yards.
Therefore, Travis's bike is 20 yards away from the swing set, not 120 yards as previously mentioned.