A particle undergoes three displacements.
The first has a magnitude of 15 m and makes
an angle of 40 ◦ with the positive x axis. The
second has a magnitude of 6.2 m and makes
an angle of 158 ◦ 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.
Edit: Part 2
Find the angle of the third displacement.
Why did the particle go on an adventure if it was just going to end up back where it started? Talk about an unnecessary trip! Anyway, to find the magnitude of the third displacement, let's call it "x". We know that the sum of all the displacements should be zero since the particle returns to its initial position. So, we can set up an equation:
15 cos(40°) + 6.2 cos(158°) + x cos(θ) = 0
Now, let's solve for x. But wait, I need some help. What's the value of θ for the third displacement?
To find the magnitude of the third displacement, we need to use vector addition.
We can break down the first displacement into its x and y components:
- The x component is given by 15 * cos(40°).
- The y component is given by 15 * sin(40°).
Similarly, we can break down the second displacement into its x and y components:
- The x component is given by 6.2 * cos(158°).
- The y component is given by 6.2 * sin(158°).
Now, we can find the resultant displacement by adding the x components and the y components separately:
- The x component of the resultant displacement is (15 * cos(40°)) + (6.2 * cos(158°)).
- The y component of the resultant displacement is (15 * sin(40°)) + (6.2 * sin(158°)).
To find the magnitude of the resultant displacement, we can use the Pythagorean theorem:
Magnitude = sqrt((x component)^2 + (y component)^2).
Let's calculate it step by step.
x1 = 15 * cos(40°)
y1 = 15 * sin(40°)
x2 = 6.2 * cos(158°)
y2 = 6.2 * sin(158°)
x_component = x1 + x2
y_component = y1 + y2
Magnitude = sqrt((x_component)^2 + (y_component)^2)
Let's calculate it.
To find the magnitude of the third displacement, we need to combine the three displacements using vector addition.
Let's call the first displacement vector A and express it as A = 15 m at 40 degrees with the positive x-axis.
Similarly, let's call the second displacement vector B, expressed as B = 6.2 m at 158 degrees with the positive x-axis.
Since the particle returns to its initial position after the third displacement, the sum of the three displacements should be zero.
To find the third displacement vector, we can subtract the sum of the first two displacements from zero.
- Sum of the first two displacements = -(A + B)
To perform vector addition, we need to break down the displacements into their x and y components. We can use trigonometry to do this.
The x-component of vector A is given by: Ax = A cos(theta), where theta is the angle it makes with the positive x-axis.
Ax = 15 m * cos(40 deg)
The y-component of vector A is given by: Ay = A sin(theta), where theta is the angle it makes with the positive x-axis.
Ay = 15 m * sin(40 deg)
Using the same process, we can find the x and y components of vector B:
Bx = 6.2 m * cos(158 deg)
By = 6.2 m * sin(158 deg)
Now, we can calculate the x and y components of the sum of the first two displacements:
Sx = Ax + Bx
Sy = Ay + By
Since the sum of all three displacements should be zero, the x and y components of the third displacement vector are given by:
Tx = -Sx
Ty = -Sy
Finally, to find the magnitude of the third displacement vector, we can use the Pythagorean theorem:
Magnitude of the third displacement vector (T) = sqrt(Tx^2 + Ty^2)
By substituting the values we calculated earlier, we can find the magnitude of the third displacement vector.