A particle undergoes two displacements, measured
from the positive x-axis, with counterclockwise
positive. The first has a magnitude
of 11 m and makes an angle of 67 ◦ with the
positive x axis. The resultant displacement
has a magnitude of 7.7 m directed at an angle
of 98 ◦
from the positive x axis.
Find the angle of the second displacement
(measured from the positive x axis, with counterclockwise
positive and within the limits of
−180◦
to +180◦
).
Answer in units of ◦
you want (x,y) such that
(11cos67,11sin67) + (x,y) = (7.7cos98,7.7sin98)
Evaluate x and y, and then find θ such that tanθ = y/x
x-comp -5.3696
y-comp -2.5
magnitude 5.92
angle 335 degrees
Well, it seems like the particle has been doing some funky moves! Let me try to help you out with this question.
We can start by breaking down the problem into two separate displacements: the first displacement with a magnitude of 11 m and an angle of 67 degrees, and the resultant displacement with a magnitude of 7.7 m and an angle of 98 degrees.
To find the angle of the second displacement, we need to subtract the angle of the first displacement from the angle of the resultant displacement. So, let's do some math:
Angle of the second displacement = Angle of the resultant displacement - Angle of the first displacement
Angle of the second displacement = 98 degrees - 67 degrees
Angle of the second displacement = 31 degrees
So, the angle of the second displacement is 31 degrees.
Remember to always keep it funny while solving problems! Let me know if there's anything else I can assist you with.
To find the angle of the second displacement, we need to determine the angle between the positive x-axis and the resultant displacement.
Let's break down the first displacement into its x and y components.
The x-component of the first displacement is given by:
x₁ = magnitude of the first displacement * cos(angle of the first displacement)
x₁ = 11 m * cos(67°)
Similarly, the y-component of the first displacement is given by:
y₁ = magnitude of the first displacement * sin(angle of the first displacement)
y₁ = 11 m * sin(67°)
Now, let's break down the resultant displacement into its x and y components.
The x-component of the resultant displacement is given by:
xᵣ = magnitude of the resultant displacement * cos(angle of the resultant displacement)
xᵣ = 7.7 m * cos(98°)
Similarly, the y-component of the resultant displacement is given by:
yᵣ = magnitude of the resultant displacement * sin(angle of the resultant displacement)
yᵣ = 7.7 m * sin(98°)
Since the x-values and y-values are additive, we can summarize the individual x and y components as follows:
x = x₁ + xᵣ
y = y₁ + yᵣ
Next, we can use the inverse tangent function to find the angle between the positive x-axis and the resultant displacement:
angle = arctan(y/x)
Substituting the x and y values we obtained earlier into the above equation, we can calculate the angle of the second displacement.
To find the angle of the second displacement, we can use vector addition. The resultant displacement is the vector sum of the two displacements.
Let's break down the given information:
First displacement:
Magnitude: 11 m
Angle with the positive x-axis: 67° counterclockwise
Resultant displacement:
Magnitude: 7.7 m
Angle with the positive x-axis: 98° counterclockwise
Now, we can use the following formula to find the angle of the second displacement:
θ = θ_r - θ_1
where θ is the angle of the second displacement, θ_r is the angle of the resultant displacement, and θ_1 is the angle of the first displacement.
Substituting the given values:
θ = 98° - 67°
Calculating:
θ = 31°
Therefore, the angle of the second displacement is 31° counterclockwise from the positive x-axis within the limits of -180° to +180°.