A particle undergoes three displacements. The first has a magnitude of 15 m and
ID: 2025454 • Letter: A
Question
A particle undergoes three displacements.
The first has a magnitude of 15 m and makes
an angle of 27 degrees with the positive x axis. The
second has a magnitude of 6.1 m and makes
an angle of 142 degrees with the positive x axis. (see
the figure below).
After the third displacement the particle
Find the angle of the third displacement (mea-
sured from the positive x axis, with counter-
clockwise positive within the limits of 180 degrees
to +180 degrees).
Answer in units of degrees.
Explanation / Answer
First displacement s = 15 cos 27 * i + 15 sin 27 j = 13.36 i +6.809 j Second displacmenet s ' =6.1 cos 142 i + 6.1 sin 142 j = -4.806 i + 3.755 j Here after third displacement it comes to initial position Net displacment S = 0 we know S = s + s' + s" From this third displacment s " = S - ( s+ s') = 0 - [ (13.36 i +6.809 j )+(-4.806 i + 3.755 j) ] = -8.554 i -10.564 j magnitude of third displacment s " = [(-8.554 ) 2+(-10.564 ) 2] = 13.592 mLet the third displacment makes an angle theta with negative x-axis Then tan ( ) = 10.554 / 8.554 = 50.97 0 Angle with positive x-axis = -180 + = -180+50.970 = 129.030 Angle with positive x-axis = -180 + = -180+50.970 = 129.030
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