Question

A jogger runs 145 m in a direction 20 degrees east of north (displacement vector A) and then 105 m in a direction 35 degrees south of east (displacement vector B). Using components, determine the magnitude and direction of the resultant vector C for these two displacements.

Answer 1

Answer:

C=155.46m ∡ 29.28° North of East

Step-by-step explanation:

The very first thing we need to do is find the components of each vector. These components are relative to the position of the angle. (Refer to the diagram).

The relations between the angles on a right triangle and the sides of the triangle are:

$sin \theta =\frac{opposite}{hypotenuse}$

$cos \theta =\frac{adjascent}{hypotenuse}$

$tan \theta =\frac{opposite}{adjascent}$

So we can use them to find the components of vectors A and B

$A_{x}=Asin\theta$

$A_{x}=(145m)sin(20^{o})=49.59m$

and likewise we can find the y-component of vector A:

$A_{y}=Acos\theta$

$A_{y}=(145m)cos(20^{o})=136.26m$

We can do the same for vector B

$B_{x}=Bcos\theta$  (notice I'm using cos this time, it's because of the position of the angle relative to the vector)

$B_{x}=(105m)cos(35^{o})=86.01m$

$B_{y}=Bsin\theta$

$B_{y}=(105m)sin(35^{o})=-60.23m$

Notice how this time the y-component of vector B is negative. This negative indicates the direction in which the vector is going. Since the y-component of vector B is going downwards, it's interpreted as a negative vector. Supposing we set the upwards direction as positive and the right direction as positive as well.

So now we can add the components to get the resultant vector C.

$C_{x}=A_{x}+B_{x}$

$C_{x}=49.59m+86.01m$

$C_{x}=135.60m$

and we can do the same with the y-components.

$C_{y}=A_{y}+B_{y}$

$C_{y}=136.26m - 60.23m$

$C_{y}=76.03m$

So now that we got the components of the resultant vector C, we can use them to find the magnitude and direction of this vector. The magnitude is found by using Pythagorean equation.

$|C|=\sqrt{C_{x}^{2}+C_{y}^{2}}$

$|C|=\sqrt{(135.6m)^{2}+(76.03m)^{2}}$

|C|=155.46m

and the angle of the vector can be found by using the inverse of tan, like this:

∡=$tan^-1(\frac{C_{y}}{C_{x}})$

∡=$tan^-1(\frac{76.03m}{135.6m})$

∡ = 29.28° North of East

So the resultant vector is:

C= 155.46m ∡ = 29.28° North of East

Answer 2

We want to find the total displacement vector given that we know two parts of the displacement.

The magnitude is 155.5m and the direction is 29.3° north of east.

Let's see how to get the answer:

First, the jogger runs 145m in 20° east of north (or 70° north of east).

If we define the north as the positive y-axis and east as the positive x-axis, we can write this vector as:

A = < 145m*cos(70°), 145m*sin(70°) > = < 49.6 m, 136.3m>

Then the jogger runs 105m in 35° south of east (or -35° north of east) this time the vector will be:

B = < 105m*cos(-35°), 105m*sin(-35°) > = <86m , -60.2m>

Vector C is just the sum of A and B:

C = A + B = < 49.6 m, 136.3m> +  <86m , -60.2m> = < 135.6m, 76.1m >

C = < 135.6m, 76.1m >

The magnitude of C is just:

|C| = √( (135.6m)^2 + (76.1m)^2) = 155.5m

The direction is given by:

θ = Atan(76.1m/135.6m) = 29.3°

This means 29.3° north of east.

If you want to learn more, you can read:

https://brainly.com/question/14033610