Math Problem Statement

  1. Solve for the resultant velocity of the vectors below:

A = 3.7m / s @0 degrees

B = 1.5m / s @38 degrees C = 6.1m / s @ 90 degrees

D = 2.2m / s @ 160 degrees

Solution

To solve for the resultant velocity of the given vectors, we need to break each vector into its horizontal (xx-component) and vertical (yy-component). Then, we can sum all the components to get the resultant vector in both directions, and finally calculate the magnitude and direction of the resultant vector.

Step 1: Break each vector into components

  • Vector A:

    • Magnitude: 3.7 m/s, Angle: 0°
    • Ax=3.7cos(0°)=3.7A_x = 3.7 \cos(0°) = 3.7
    • Ay=3.7sin(0°)=0A_y = 3.7 \sin(0°) = 0
  • Vector B:

    • Magnitude: 1.5 m/s, Angle: 38°
    • Bx=1.5cos(38°)B_x = 1.5 \cos(38°)
    • By=1.5sin(38°)B_y = 1.5 \sin(38°)
  • Vector C:

    • Magnitude: 6.1 m/s, Angle: 90°
    • Cx=6.1cos(90°)=0C_x = 6.1 \cos(90°) = 0
    • Cy=6.1sin(90°)=6.1C_y = 6.1 \sin(90°) = 6.1
  • Vector D:

    • Magnitude: 2.2 m/s, Angle: 160°
    • Dx=2.2cos(160°)D_x = 2.2 \cos(160°)
    • Dy=2.2sin(160°)D_y = 2.2 \sin(160°)

Now, I'll calculate the exact values for BxB_x, ByB_y, DxD_x, and DyD_y.

Step 2: Calculate the components

Let me calculate those.The components of each vector are as follows:

  • Vector A:

    • Ax=3.7A_x = 3.7
    • Ay=0.0A_y = 0.0
  • Vector B:

    • Bx=1.18B_x = 1.18
    • By=0.92B_y = 0.92
  • Vector C:

    • Cx=0.0C_x = 0.0
    • Cy=6.1C_y = 6.1
  • Vector D:

    • Dx=2.07D_x = -2.07
    • Dy=0.75D_y = 0.75

Step 3: Sum the components

To get the resultant vector's components, we add the individual components:

Rx=Ax+Bx+Cx+DxR_x = A_x + B_x + C_x + D_x Ry=Ay+By+Cy+DyR_y = A_y + B_y + C_y + D_y

Now, let's compute the resultant components.The resultant components are:

  • Rx=2.81R_x = 2.81
  • Ry=7.78R_y = 7.78

Step 4: Calculate the magnitude and direction of the resultant vector

  • The magnitude RR is given by: R=Rx2+Ry2R = \sqrt{R_x^2 + R_y^2}

  • The direction θ\theta (relative to the positive xx-axis) is: θ=tan1(RyRx)\theta = \tan^{-1} \left( \frac{R_y}{R_x} \right)

Let's compute the magnitude and direction.The resultant velocity is:

  • Magnitude: 8.27 m/s
  • Direction: 70.10° relative to the positive xx-axis.

Would you like more details or have any questions?

Related questions:

  1. How do you break a vector into components?
  2. Why is the angle measured from the positive xx-axis?
  3. What are some common applications of vector addition in physics?
  4. How can you add more than two vectors graphically?
  5. What is the difference between velocity and speed?

Tip:

Always draw a diagram when working with vectors to visualize the direction and components clearly.

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Math Problem Analysis

Mathematical Concepts

Vector Addition
Trigonometry
Component Method for Vectors

Formulas

Vector components: V_x = V * cos(theta), V_y = V * sin(theta)
Resultant vector magnitude: R = sqrt(R_x^2 + R_y^2)
Resultant vector direction: theta = tan^(-1)(R_y / R_x)

Theorems

Pythagorean Theorem
Trigonometric Ratios

Suitable Grade Level

Grades 10-12