Aman pushing a mop across a floor causes it to undergo two displacements. the first has a magnitude of 156 cm and makes an angle of 126° with the positive x axis. the resultant displacement has a magnitude of 133 cm and is directed at an angle of 34.0° to the positive x axis. find the magnitude and direction (counterclockwise of the positive x axis) of the second displacement.

second displacement is  R = 208.6 cm  and  θ = 346º

Explanation:

This is a problem of adding vectors, the easiest way to work these problems is to decompose the vectors and find the resulting vectors on each axis

Let's use trigonometry to break down each displacement vector. Let's start with vector 1 that has a magnitude m1 = 156 cm

          sin 126 = Y1 / m1

          Y1. = m1 without 126

          Y1 = 156 without 126

          Y1 = 126.2 cm

          cos 126 = X1 / m1

          X1 = m1 cos 126

          X1 = 156 cos 126

          X1 = - 91.69 cm

The resulting vector has R = 133 cm

         sin 34 = Ry / R

         Ry = R sin 34

         Ry = 133 sin 34

         Ry = 74.37 cm

         cos 34 = Rx / R

         Rx = R cos 34

         Rx = 133 cos 34

         Rx = 110.3 cm

We already have all the components, we can add algebraically on each axis

X axis

         Rx = X1 + X2

         X2 = Rx -X1

         X2 = 110.3 - (-91.69)

         X2 = 202 cm

Y Axis  

         Ry = Y1 + Y2

         Y2 = Ry - Y1

         Y2 = 74.37 -126.3

         Y2 = -52 cm

Let's build the resulting vector

          R = (202 i ^ + 52 y ^) cm

          R = (202, -52) cm

We can also use the Pythagorean triangle and trigonometry to find the module and direction

          R² = Rx² + Ry²

          R = √(202² + 52²)

          R = 208.6 cm

          tan θ = Ry / Rx

          tan θ = -52/202

          θ = tan⁻¹ (-0.257)

          θ = -14.4º

The negative sign indicates that it is measured from the x-axis clockwise, to measure counterclockwise from the x-axis

          θ = 360-14

           θ = 346º

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

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