A jogger ran 3 miles due east of his house. then he ran 5 miles at a heading of 30o east of north (or 30o ne). how far is he fro

Question

3 years ago

5

m his house after running 8 miles

1 answer:

Natasha_Volkova [10] · Answer 1 · 2022-01-30T04:00:56+03:00

Magnitude of the net displacement: 7.0 miles

Step-by-step explanation:

In order to find the total displacement of the man, we have to resolve each displacement along the x- and y- direction.

Taking x as positive x direction and y as positive north direction:

- The first motion is 3 miles due east, so

$A_x = 3 mi\\A_y = 0$

- The second motion is 5 miles at $30^{\circ}$ east of north, so:

$B_x = 5 sin 30^{\circ}=2.5 mi\\B_y = 5 cos 30^{\circ}=4.3 mi$

So the components of the net displacement are

$R_x = A_x + B_x = 3 + 2.5 = 5.5 mi\\R_y = A_y+B_y = 0 + 4.3 = 4.3 mi$

And so the magnitude of the net displacement is

$d=\sqrt{R_x^2+R_y^2}=\sqrt{5.5^2+4.3^2}=7.0 mi$

So, he is 7.0 miles far from the house.

Learn more about distance and displacement:

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