Vector V1 is 6.4 units long and points along the negative x axis. Vector V2 is 8.9 units long and points at 60 degrees to the positive x axis
1) find x and y components of v1 and v2?
2) find magnitude of the sum v1+v2?
3) find the angle of the sum v1+v2?
I found the answers to part 1 (v1= -6.4,0 and v2= 4.5,7.7) and part 2 (8.0) but need help with part 3. thanks!
I assume 60 degrees in the +y direction
V1x = -6.4
V1y = 0
V2x = 8.9 cos 60 = 4.45 agree
V2y = 8.9 sin 60 = 7.71 agree
Vx= -6.4 + 4.45 = -1.95
Vy= 0+ 7.71 = 7.71
sqrt(Vx^2+Vy^2) = 7.95 agree
tangent of angle above -x axis = 7.71/1.95
= 3.954
so that angle = 75.8 degrees above -x axis
which is
180 - 75.8 = 104.2 degrees counterclockwise from +x axis
ALWAYS sketch a graph !!!!
Sure, I'd be happy to help with part 3! To find the angle of the sum V1 + V2, we can use the inverse tangent function (arctan).
1) The x-component of V1 is -6.4 and the x-component of V2 is 4.5.
2) The y-component of V1 is 0 and the y-component of V2 is 7.7.
Now, let's find the x and y components of the sum V1 + V2:
The x-component of the sum = (-6.4) + 4.5 = -1.9
The y-component of the sum = 0 + 7.7 = 7.7
To find the angle, we can use the arctan function:
angle = arctan(y-component of the sum / x-component of the sum)
angle = arctan(7.7 / -1.9)
Now let's bust out our calculators:
angle ≈ -76.7 degrees
Therefore, the angle of the sum V1 + V2 is approximately -76.7 degrees. Keep in mind that this angle is relative to the positive x-axis.
To find the angle of the sum of vectors V1 and V2, you can use the cosine rule. The cosine rule states that:
c^2 = a^2 + b^2 - 2ab * cos(C)
where c is the magnitude of the sum of vectors V1 and V2, a is the magnitude of V1, b is the magnitude of V2, and C is the angle between the vectors.
In this case, we have the magnitudes of V1 and V2 as 6.4 and 8.9 respectively. We also know the magnitude of the sum V1 + V2 is 8.0 (as you found out).
Now, we can rearrange the cosine rule to solve for the angle C:
cos(C) = (a^2 + b^2 - c^2) / (2ab)
Substituting the values, we have:
cos(C) = (6.4^2 + 8.9^2 - 8.0^2) / (2 * 6.4 * 8.9)
cos(C) = (40.96 + 79.21 - 64) / (112.96)
cos(C) = 56.17 / 112.96
cos(C) ≈ 0.4979
Now, to find the angle C, we can take the inverse cosine (cos^-1) of 0.4979:
C ≈ cos^-1(0.4979)
C ≈ 60.14 degrees
Therefore, the angle of the sum V1 + V2 is approximately 60.14 degrees.
To find the angle of the sum of v1 and v2 (part 3), you can use trigonometry. Follow these steps:
1. Find the x and y components of the sum of v1 and v2:
- The x component (let's call it Vx_sum) is the sum of the x components of v1 and v2: Vx_sum = Vx1 + Vx2.
- The y component (let's call it Vy_sum) is the sum of the y components of v1 and v2: Vy_sum = Vy1 + Vy2.
2. Calculate the magnitude of the sum using the Pythagorean theorem:
- The magnitude (let's call it V_sum) of the sum is given by: V_sum = sqrt(Vx_sum^2 + Vy_sum^2).
3. Find the angle (let's call it theta_sum) using trigonometry:
- The angle theta_sum can be calculated using the inverse tangent function: theta_sum = atan(Vy_sum / Vx_sum).
Now let's solve part 3 using the values you provided:
Vx1 = -6.4, Vy1 = 0
Vx2 = 4.5, Vy2 = 7.7
1. Calculate the x and y components of the sum:
Vx_sum = Vx1 + Vx2 = -6.4 + 4.5 = -1.9
Vy_sum = Vy1 + Vy2 = 0 + 7.7 = 7.7
2. Calculate the magnitude of the sum:
V_sum = sqrt(Vx_sum^2 + Vy_sum^2) = sqrt((-1.9)^2 + 7.7^2) = sqrt(3.61 + 59.29) = sqrt(63.9) = 8.0
3. Find the angle of the sum:
theta_sum = atan(Vy_sum / Vx_sum) = atan(7.7 / -1.9) = atan(-4.052631578947368) ≈ -68.14 degrees
Therefore, the angle of the sum of v1 and v2 is approximately -68.14 degrees.