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How to Calculate Bearing in Physics

Understanding How to Calculate Bearing in Physics

In this post, I will guide you on how to calculate bearing in physics. But we need to first understand the definition of bearing.

Definition: Bearing is the angular dispalcement or direction between cardinal points in an x – y plane.

To find the direction of an object, we need to have a look at a plane or space. By looking at the four cardinal points from the perspective of a plane, it is obvious that we have x and y cordinates and we call them abcissa and ordinate respectively.

Additionally, we assign names to these four cardinal points or poles, and we call them north, south, west, and east.

While calculating bearing,we use the symbol of θ to indicate the angular direction of an object. Here are some the methods we apply to measure angular direction (θ):

  • Applying compass
  • Using protractor on a map
  • Simple calculation

In this article, I will solve few examples that will guide you on how to calculate bearing in physics.

Examples of How to Calculate Bearing in Physics

We have four cardinal points starting from North (N) to East (E), from N to E covers an angle of 900. Depending on the direction of the object, the angle between two cardinal points is 900.

This makes movement from North to East, and then to South 1800

From North to East to South and then to West 2700.

And complete movement from North, East, South, West, and back to North 3600

After finding the direction of the object, this is how you write it:

Assuming your θ is 600, you now know that the direction θ=600 is between North and East

Thus, you write it as N 600E

Which means sixty degrees from North direction towards East

Now, here are few examples to guide you understand how to calculate bearing in physics

Example 1

A man walks 4 kilometer due Northand then walks 3km due East. What is his displacement from the starting point.


How to Calculate Bearing in Physics


We first extract our data from the above question

Let us assume that his initial position is the origin O

He walks from his origin O to another point A due North = 4 kilometer

Additionally, He walks from his second position A to another location B due East = 3 kilometer

For us to find his displacement from the starting point, we calculate OB.

To calculate OB, we apply the formula below

OB2 = OA2 + AB2

We have OB = ?, OA = 4km, and AB = 3km

Substituting the above values into the formula, we have

OB2 = 42 + 32 = 16 + 9 = 25

Therefore we take the square root of both sides, to find OB

OB = √25 = 5km

Now we can see that the dispalcement of the man from his initial position is 5km.

We need to also find the direction, θ

By applying the formula below, we have

tan θ = AB/OA = 3/4

Devide both sides by tan to find θ

θ = tan-1 (3/4) = tan-1 (0.75) = 36.90

We can now approximate θ as 370 which is the direction of the man

θ = 370

Because 370 is between North(N) and East (E), we can now say that the direction of the man is N 370E ( The man is thirty seven degrees from North direction towards East)

Therefore, our final answer is as follows:

The displacement is 5km, N 370E

Example 2

Find the location of A(3,-6) and B(7,9) with the reference to a point (0,0) and determine the bearing of A,



Let us consider triangle OAB

When we trace our line on B axis from Origin(which is zero) to be on A(3,-6), we can see that X1 = 3 and this implies that OB = X1 = 3

Therefore, modulus of OB= 3 units

modulus of AB= 6 units

We now apply the formula below

tan θ = OB/AB = 3/6

By making θ subject of the formula, we have

θ = tan-1 (0.5) = 26.60

Which can be approximated into

θ = 270

To find the bearing of A, we add 270 to the 900 which si being completed from North to East

Therefore, A = 900 + 270 = 1170

Which is considered as E 1170S

Example 3

A butterfly moves from a point X to a point Y, a distance of 15 meters in the direction 0450. It then changes direction and moves to a point Z, in the direction 1550. If the point Z is due east of X, find the distance of the point Z from the point Y.



By applying the sine rule which is

X / Sinθ1 = Y / Sinθ2 = Z / Sinθ3

and θ1 = 450, θ2 = 700, θ3 = 650, Z = 15m, and X = ?

we now substitute the above data into X / Sinθ1 = Z / Sinθ3

X / Sin450 = 15 / Sin650

We can now cross multiply the above equation

XSin650 = 15 x Sin450

We now make X subject of the formula

X = (15 x Sin450) / (Sin650) = (15 x 0.71) / (0.91) = 10.65/0.91 = 11.7m

Therefore the distance of the point Z from the point Y which is X = 11.7 meters

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