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DochEvi [55]
3 years ago
5

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

m his house after running 8 miles
Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
4 0

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:

brainly.com/question/3969582

#LearnwithBrainly

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