McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
7. Vectors
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Exercise 42 Page 689

Practice makes perfect
a We are given that a runner's velocity is 6 miles per hour due east and the wind is blowing 2 miles per hour due north, and we're asked to draw a diagram describing the given situation. Let's start with drawing arrows describing the given velocities.

As we can see, the component form of the vector representing the velocity of the runner is ⟨ 6,0⟩, and the component form of the vector representing the velocity of the wind is ⟨ 0, 2 ⟩. Therefore, the resultant vector is the sum of these two vectors. ⟨ 6,0 ⟩ + ⟨ 0, 2 ⟩ = ⟨ 6, 2⟩ The vector ⟨ 6,2 ⟩ represents the resultant velocity, which we will call r.

b We are asked to determine the resultant velocity of the runner, which means we need to find the resultant speed and the resultant position of r. To find the resultant speed, the magnitude of r, we will use the Distance Formula. Let's substitute ( , ) for the initial point and ( 6, 2) for the terminal point.
| r|=sqrt((x_2-x_1)^2+(y_2-y_1)^2)
|r|=sqrt(( 6- )^2+( 2- )^2)
Simplify right-hand side
|r|=sqrt(6^2+2^2)
|r|=sqrt(36+4)
|r|=sqrt(40)
|r|=6.3245...
|r|≈ 6.3
The resultant speed is approximately 6.3 miles per hour. To evaluate the resultant direction, we need to find the measure of an angle that vector r forms with a north-south line. We will call this angle α.

If we call the angle between the red and blue arrows θ, then we can notice that α and θ add to be 90^(∘).

To find the measure of θ, we can use one of the trigonometric ratios as the vectors form a right triangle. Let's recall that in a right triangle the tangent of an angle is the ratio between the leg opposite to this angle and leg adjacent to this angle. Using this definition, we can write an equation. tan θ=2/6 Next, we can rewrite the equation using the inverse tangent to evaluate the measure of θ. tan θ=2/6 ⇓ θ=tan ^(-1)2/6 ≈ 18.4^(∘) Knowing the measure of θ, we can find the measure of α.
α+θ=90^(∘)
α+18.4^(∘)=90^(∘)
α=71.6^(∘)
The resultant direction is approximately 71.6^(∘) east of north. Therefore, the runner's resultant velocity is approximately 6.3 miles per hour at the angle of 71.6^(∘) east of north.