Question

A particle undergoes two displacements. The first has a

magnitude of 150 cm and makes an angle of 120° with
the positive x axis. The resultant displacement has a magnitude
of 140 cm and is directed at an angle of 35.0° to
the positive x axis. Find the magnitude and direction of
the second displacement.

Answer 1

To determine the second displacement, vector addition is used to subtract the first vector's components from the resultant vector's components. This allows for the calculation of the magnitude and direction of the second displacement.

Step-by-step explanation:

To find the magnitude and direction of the second displacement, we need to use vector addition principles. The first displacement vector δ1 has a magnitude of 150 cm and direction of 120° from the positive x-axis, which can be broken down into horizontal (x) and vertical (y) components using trigonometry:

• x-component: 150 cm * cos(120°)
• y-component: 150 cm * sin(120°)

The resultant displacement vector δ has a magnitude of 140 cm at 35° to the positive x-axis, also with x and y components:

• x-component: 140 cm * cos(35°)
• y-component: 140 cm * sin(35°)

To find the second displacement vector δ2, we use the fact that δ = δ1 + δ2. This implies that δ2 = δ - δ1. We subtract the components of δ1 from the components of δ to find the components of δ2. After calculating the components, we use the Pythagorean theorem to find the magnitude of δ2 and inverse tangent function to find its direction.

Answer 2

First you need to draw the picture of the problem to better understand it. Like the one bellow.

In this task you have 2 sides of triangle and we can calculate angle between them. Angle between them is 120 - 35 = 85 degrees.

Once you have those 3 variables you can calculate third side of triangle using cosine law.
a - second displacement
b - first displacement
c- resultant displacement.


`a^(2) = b^(2)+ c^(2)-2*b*c*cos85`
now we just need to calculate this.

`a^(2) = 38439.458`
a = 196

now, we use cosine law again to find the angle between second and first displacement.

`\alpha = 45.36` degrees

The angle marked with "?" in the graph is our direction angle. We will call it
`\beta`


`\beta =90-30-45.36 = 14.64`

Second displacement has magnitude of 196 and a direction of -14.64 with positive x axis