Question

In this question, i is a unit vector due east and j is a unit vector due north. A cyclist rides at a speed of 4 m/s on a bearing of 015 degrees. a) Write the velocity vector of the cyclist in the form of xi + yj, where x and y are constants. b) A vector of magnitude 6 on a bearing of 300 degrees is added to a vector of magnitude 2 on a bearing of 230 degrees to give vector v. Find the magnitude and bearing of v.

Answer 1

Answer:

a) 1.04i + 3.86j

b) magnitude = 8; bearing = 302.7°

Step-by-step explanation:

a)

The  first diagram represents the velocity vector of the cyclist.

To express this vector in the form xi + yj, we have to find the components of the vector in the horizontal (i) and vertical (j) directions.

If we consider the horizontal component of the vector to be $x$, and the vertical component to be $y$, then:

• horizontal component ⇒ $sin (15^{\circ}) = \frac{x}{4}$

                                       ⇒ $x = 4\space\ sin(15^{\circ})$

                                       ⇒ $x \approx \bf 1.04$

• vertical component ⇒ $cos(15^{\circ}) = \frac{y}{4}$

                                   ⇒ $y = 4 \space\ cos(15^{\circ})$

                                   ⇒ $y \approx \bf 3.86$

Now that we have the values of both the horizontal and vertical component, we can write the vector in the form of xi + yj:

vector ⇒ 1.04i + 3.86j

b)

The second diagram shows the first vector (red), the second vector (blue), and the resultant vector v (black). The dashed lines represent the components of the respective vectors.

To add two vectors given their magnitudes and direction, we have to add their components.

In order to find the horizontal and vertical components of the given vectors, we can use a method similar to that used above, so that:

○ For the first vector (magnitude 6):

• horizontal component ⇒ $x = 6 \space\ sin (60^{\circ})$

                                       ⇒ $\bf 5.2$

• vertical component ⇒ $y = 6 \space\ cos(60^{\circ})$

                                   ⇒ $y = \bf 3$

○ For the second vector (magnitude 2):

• horizontal component ⇒ $x = 2 \space\ cos (40^{\circ})$

                                       ⇒ $\bf 1.5$

• vertical component ⇒ $y = 2 \space\ sin(40^{\circ})$

                                   ⇒ $\bf 1.3$

Now we can add the respective components together:

v = 5.2i + 3j  +  1.5i + 1.3j

 ⇒ (5.2 + 1.5)i + (3 + 1.3)j

 ⇒  6.7i + 4.3j

∴ Magnitude of v ⇒ $|v| = \sqrt{(6.7)^2 + (4.3)^2}$

                             ⇒ $|v| \approx \bf 8$

To find the bearing of v, we have to first calculate the angle marked $\alpha$:

$tan \alpha = \frac{4.3}{6.7}$

⇒ $\alpha = tan^{-1}(\frac{4.3}{6.7})$

⇒ $\alpha = \bf 32.7^{\circ}$

∴ Bearing = 270° + 32.7°

                = 302.7°