A student bikes to school by traveling first dN = 0.800 miles north, then dW = 0

Question

A student bikes to school by traveling first dN = 0.800 miles north, then dW = 0.300 miles west, and finally dS = 0.100 miles south.

Take the north direction as the positive y direction and east as positive x. The origin is still where the student starts biking. Let d N be the displacement vector corresponding to the first leg of the student's trip. Express d N in component form.

Express your answer as two numbers separated by a comma (e.g., 1.0,2.0). By convention, the x component is written first.

Explanation / Answer

here

the first leg

dNx = 0; dNy = + 0.8 mi ......................(0 , 0.8)

the second leg

dWx = - 0.3 mi; dWy = 0 ...................(-0.3 , 0)

the third leg

dSx = 0; dSy = - 0.1 mi ......................(0 , -0.1)

then the displacement vector is

dN = ( -0.3 , (0.8 - 0.1) )

dN = ( -0.3 , 0.7)

then the magnitude is = sqrt( 0.3^2 + 0.7^2 )

= 0.76

then the direction is

theta = tan^-1( 0.7 / 0.3 ) = 66.8 deg

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