A particle undergoes three displacements.

The first has a magnitude of 15 m and makes an angle of 30 degrees with the positive x axis. The second has a magnitude of 8.6 m and makes an angle of 149� degrees with the positive x axis. After the third displacement the particle returns to its initial position.
Find the magnitude of the third displacement.
Answer in units of m

Why did the particle go on a vacation and then come back to its initial position? Maybe it just wanted a quick getaway from all the physics equations! But don't worry, I'll help you find the magnitude of the third displacement.

To find the magnitude of the third displacement, we need to consider the vector sum of the first two displacements.

Let's call the first displacement D1 and the second displacement D2.

D1 has a magnitude of 15 m and an angle of 30 degrees with the positive x-axis.
D2 has a magnitude of 8.6 m and an angle of 149 degrees with the positive x-axis.

To find the third displacement, we can add D1 and D2 together.

But hold on, before we add them up, we need to make sure they're in the same coordinate system. Let's convert D2's angle to be in the range from 0 to 360 degrees. Adding 360 degrees to 149 degrees gives us 509 degrees.

Now, let's add D1 and D2 together:

D = D1 + D2

To do this, we'll break D1 and D2 into their x and y components:

D1x = 15 m * cos(30 degrees)
D1y = 15 m * sin(30 degrees)

D2x = 8.6 m * cos(509 degrees)
D2y = 8.6 m * sin(509 degrees)

Now, let's add the x-components and the y-components separately:

Dx = D1x + D2x
Dy = D1y + D2y

Finally, we can find the magnitude of the third displacement using the Pythagorean theorem:

D = sqrt(Dx^2 + Dy^2)

After all that math, the magnitude of the third displacement is... wait for it... *drumroll*... the answer in units of meters!

To find the magnitude of the third displacement, we can use the concept of vector addition.

Let's break down the first two displacements into their x and y components.

For the first displacement:
Magnitude = 15 m
Angle = 30 degrees

The x component can be found using cosine:
x1 = 15 m * cos(30 degrees) = 15 m * (sqrt(3)/2) ≈ 12.99 m

The y component can be found using sine:
y1 = 15 m * sin(30 degrees) = 15 m * (1/2) = 7.5 m

So, the first displacement can be represented by the vector (12.99 m, 7.5 m).

For the second displacement:
Magnitude = 8.6 m
Angle = 149 degrees

The x component can be found using cosine:
x2 = 8.6 m * cos(149 degrees) = 8.6 m * (-0.6235) ≈ -5.36 m

The y component can be found using sine:
y2 = 8.6 m * sin(149 degrees) = 8.6 m * (-0.7818) ≈ -6.7 m

So, the second displacement can be represented by the vector (-5.36 m, -6.7 m).

To find the third displacement, we need to add the first two displacements:

x3 = x1 + x2 = 12.99 m + (-5.36 m) ≈ 7.63 m
y3 = y1 + y2 = 7.5 m + (-6.7 m) ≈ 0.8 m

So, the third displacement can be represented by the vector (7.63 m, 0.8 m).

To find the magnitude of the third displacement, we can use the Pythagorean theorem:
Magnitude = sqrt((7.63 m)^2 + (0.8 m)^2)
= sqrt(58.44 m^2 + 0.64 m^2)
= sqrt(59.08 m^2)
≈ 7.68 m

Therefore, the magnitude of the third displacement is approximately 7.68 m.

To find the magnitude of the third displacement, we need to consider the vector addition of the first two displacements.

First, let's determine the x and y components of each displacement:

For the first displacement:
Magnitude = 15 m and angle = 30 degrees
x component = magnitude * cos(angle) = 15 m * cos(30°) = 12.99 m
y component = magnitude * sin(angle) = 15 m * sin(30°) = 7.5 m

For the second displacement:
Magnitude = 8.6 m and angle = 149 degrees
x component = magnitude * cos(angle) = 8.6 m * cos(149°) = -8.19 m
y component = magnitude * sin(angle) = 8.6 m * sin(149°) = 4.07 m

Next, we can find the resultant x and y components by adding the corresponding components from the first and second displacements:

Resultant x component = 12.99 m + (-8.19 m) = 4.8 m
Resultant y component = 7.5 m + 4.07 m = 11.57 m

Now we can use these components to find the magnitude of the third displacement:

Third displacement magnitude = sqrt((Resultant x component)^2 + (Resultant y component)^2)
= sqrt((4.8 m)^2 + (11.57 m)^2)
= sqrt(23.04 m^2 + 134.2449 m^2) = sqrt(157.2849 m^2)
= 12.53 m

Therefore, the magnitude of the third displacement is approximately 12.53 m.

convert the two displacements to (x,y) values

add the two x & y values
that will be the final displacement. magnitude is √(x^2+y^2)