Answer to Question #136365 in Mechanics | Relativity for Alfred John Aguilar

Question #136365
A hiker walks 2.5 km, 53.1°N of E. Then he walks another 2.0 km eastward. How far is the hiker from his starting point?
1
Expert's answer
2020-10-07T08:58:52-0400

Solution

According to question diagram can be shown as below



Components of Vector A are

Ax=2.5×cos53.10  =1.5kmA_x=2.5\times cos53.1^0\\ \space\space=1.5km

Ay=2.5×sin53.10=2kmA_y=2.5\times sin53.1^0=2 km


Components of Vector B

Bx=2×cos00=2kmB_x=2\times cos0^0=2km

By=0kmB_y=0 km


Resultant vector C can be written as

Cx=Ax+Bx=1.5+2=3.5kmC_x=A_x+B_x=1.5+2\\=3.5 km

Cy=Ay+By=2+0C_y=A_y+B_y=2+0\\ =2 km

Therefore distance from starting point

C=Cx2+Cy2=(3.5)2+22=4.03kmC= \sqrt{C_x^2+C_y^2} \\= \sqrt{(3.5)^2+2^2}\\=4.03 km

Therefore hiker is 4.03 Km from starting point.




Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment