A particle undergoes two displacements. The first has a magnitude of 11 m and ma

Question

A particle undergoes two displacements. The first has a magnitude of 11 m and makes an angle of 70 with the positive x axis. The result after the two displacements is 9 m directed at an angle of 130 to the positive x axis using counterclockwise as the positive angular direction.

Find the angle of the second displacement (measured counterclockwise from the positive x axis). Answer in units of .

Side note, I've tried solving (11cos70,11sin70) + (x,y) = (9cos130,9sin130) with arctan(y/x) = 19.86 (in degrees, I checked) yet the automated homework checker keeps telling me this is not correct.

Explanation / Answer

A = 11 m , angle a = 70°

the resultant R = 9 m, angle r = 130°

Find the magnitude B and direction of the second displacement (angle = b)

R = A + B (vector sum)

B = R - A (vector difference)

Bx = Rx - Ax = Rcos(r) - Acos(a) = 9*cos130 - 11*cos70 = -2.01 m

By = Ry - Ay = Rsin(r) - Asin(a) = 9*sin130 -11*sin70 = -3.44 m

so B =3.98 m

note Bx < 0, By < 0, so angle b is in 3rd quadrant,

angle b = 180 + arctan(3.44/2.01) = 239.68 degree

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