Concept of Displacement in Physics
In this comprehensive article, you will learn how to find displacement in physics. Displacement refers to the change in position of an object from its initial point to its final point.
Unlike distance, which only considers the length of the path travelled, displacement takes into account both the distance and direction of the movement.
Additionally, displacement is a vector quantity, meaning it has both magnitude and direction. Displacement is an essential concept in physics, especially when analyzing the motion of objects in different scenarios.
You can watch the video below to learn how to find displacement using bearing
What is Displacement in Physics?
Definition of Displacement in Physics: Displacement is the measure of separation between two points in a specified direction. Displacement is a vector quantity and is measured in meters.
Furthermore, displacement is defined as the straight-line distance between the initial and final positions of an object, along with the direction of this distance. Mathematically, it can be represented as:
Displacement (d) = rf – ri
Where:
d is the dispalcement vector
rf is the final position vector
ri is the initial position vector
When a person moves from one point to another (changes his position) in a straight, specified direction. We measure his distance as displacement. Displacement is direct, straight, and in the form of translational motion. Unlike distance, the line of movement for displacement is not in random or zig-zag motion.
To perfectly describe displacement, let us assume that a man joined a train, and the train moves from city Y to city Z in a straight line covering a distance of 300 kilometres. After reaching town Z, the train followed the same route and stopped at 80 kilometres before reaching it is destination.
To find the displacement of the man’s journey, we subtract 80km from the 300km due to the fact that the route is the same and straight. Therefore, the displacement of the man is ( 300km – 80km = 220km) 220km.
Displacement shares similarities with distance because they are both measured in meters, and they are a measure of separation between two points. However, displacement is the exact opposite of distance, this is because displacement is in a specified direction while distance is not.
Displacement is related to velocity while distance is related to speed. They are both related to time.
Displacement Formula in Physics
The displacement formula in Physics are as follows:
(1) s = v x t, (2) s =ut + 1/2 at2, (3) S = (v2 -u2 )/2a, (4) S = √[ (x2-x1)2 + (y2-y1)2 + (z2-z1)2 ], (5) xtotal = xfinal – xinitial
Explanation: Displacement Formula
- Since Speed=Distance/time, we can also say that velocity(v)=Displacement(s)/time(t). By making displacement subject of the formula, we now have Displacement = velocity x time. Therefore, we can now write s = v x t
- We can also apply equations of motion to find the displacement of an object which is S =ut + 1/2 at2 or
- S = (v2 -u2 )/2a
- We can also apply S = √[ (x2-x1)2 + (y2-y1)2 + (z2-z1)2 ]
- Additionally, we can calculate displacement by applying the formula which says xtotal = xfinal – xinitial (where xtotal= total diplacement, xinitial = initial displacement, and xfinal = final displacement)
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How to Calculate Displacement
To calculate displacement, we need to know the initial and final positions of the object. Let’s consider a few scenarios to understand how displacement is calculated.
Scenario 1: Linear Motion in One Dimension
In the case of linear motion along a single dimension, calculating displacement is straightforward. If an object moves from an initial position xi to a final position xf, the displacement Δx can be calculated as:
Δx = xf – xi
Scenario 2: Two-Dimensional Motion
When dealing with two-dimensional motion, we consider both horizontal and vertical displacements. For an object moving in the xy-plane, the displacement d can be calculated using the Pythagorean theorem:
|d| = √(Δx)2 + √(Δy)2
Where Δx is the change in the horizontal position and Δy is the change in the vertical position.
Solved Problems: How to Find Displacement in Physics
Here are some examples to help you understand how to find displacement in physics.
Problem 1
A ball was displaced from O to A along a straight line, a distance of 5-meters. It was later returned to O along the same path. What is the resultant dispalcement?
Solution
Data:
The displacement from O to A is 5m
Coming back from A to O along the same path implies that AO is also 5m
Thus,
Displacement OA = 5m
Displacement AO = 5m
The total displacement from O to A, and then back to O is the summation of OA and -AO
AO is negative because it is ccoming back to the same direction and cancelling the initial displacement
Thus,
To find the resultant displacement, we say:
The resultant displacement = OA + (-AO) = 5m + (-5m) = 5m – 5m = 0
Therefore, the total displacement from O to A and back to O is zero. This shows that there is no any displacement by the ball.
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Problem 2
Determine the distance between two points A(5,-8) and B(9,12)
Solution
Data:
After reading the above question, we can see:
x1 = 5, x2 = 9, y1 = -8, and y2 = 12
Thus, by applying the formula S = √[ (x2-x1)2 + (y2-y1)2 ]
We now have
S = √[ (9 – 5)2 + (12 – {-8})2 ]
which is equal to
S = √[ (9 – 5)2 + (12 + 8)2 ]
After summing the right-hand side (RHS), we have
S = √[ (4)2 + (20)2 ] = √ 16 + 400 = √416 = 20.4 units
Problem 3
A body starts with an initial velocity of 4 m/s and continues to move with an acceleration of 1.5m/s2. Find the displacement moved in 6-seconds
Solution
Data:
Initial velocity, u = 4m/s
acceleration, a = 1.5m/s2
time, t = 6 seconds
displacement = ?
To find the displacement, we need to obtain our average velocity by applyig the formula [ (u+v)/2 ] and v is the final velocity
Thus,
Average velocity = (Final velocity + Initial velocity)/2
But we don’t have v. Let us apply one of the equations of motion to find v which is the final velocity
v = u + at = 4 + (1.5 x 6) = 4 + 9 = 13
This implies that
Average velocity = (v+u)/2 = (4+13)/2 = 17/2 = 8.5m/s
Therefore, we can now find the displacement by applying the formula
Displacement, S = Average Velocity x Time
Thus,
Displacement = 8.5 x 6 = 51m
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Problem 4
A body accelerates uniformly from rest at the rate of 3ms-2 for 8 seconds. Calculate the distance covered by the body during the acceleration.
Solution
Data
Acceleration of the body, a = 3ms-2
The time it takes the body to accelerate, t = 8s
Because the body starts from rest, we can say that the initial velocity is zero, u = 0
Distance covered by the body during the acceleration is the displacement of the body during that acceleration. This is because acceleration is due to the effect of velocity which is also a vector quantity.
Therefore, we can now see that
Distance covered by the body during the acceleration = Displacement = ?
To find the Displacement S, we apply one of the equations of motion that says
S = ut + 1/2 (at2)
Therefore, by susbtituting our data into the above formula, we will have
S = 0 x 8 + 1/2 ( 3 x 82)
Which is equal to
S = 0 + 1/2 (3 x 64)
And the above expression will be
S = 1/2 (192)
When we multiply 1/2 by 192, we will have
S = 192/2 = 96m
Therefore, the distance covered by the body during the acceleration which is equivalent to displacement is 96 meters.
Problem 5
A bus travelling at 15m/s accelerates uniformly at 4ms-2. What is the distance covered in 10 seconds?
Solution:
Data:
Initial velocity, u = 15m/s
The acceleration of the bus, a = 4ms-2
The distance covered by the car is the displacement of the bus. This is because we are dealing with acceleration, which is due to velocity, and velocity is the ratio of displacement to time.
Thus, Distance covered by the bus = displacement, s = ?
The time taken by the car to accelerate, t = 10s
Therefore, since s = ut + 1/2(at2)
we substitute our data into the above formula
s = 15 x 10 + 1/2(4 x 102)
Which is equal to
s = 15 x 10 + 1/2(4 x 100)
s = 150 + 200 = 350m
Therefore the displacement s is 350 meters.
Problem 6
A boy travels 8 kilometres eastward to a point B and then 6-kilometre northward to another point C. Determine the difference between the magnitude of the displacement of the boy and the distance travelled by him.
Solution
Data
From the above question, we can see that
OC = ?
OB = 8km
and BC = 6km
By applying the formula OC2 = OB2 + BC2
we now have
OC2 = 82 + 62 = 64 + 36 = 100
Thus, OC2 = 100
By taking the square root of both sides, we will have
√OC2 = √100
which implies that
OC = √100 = 10km
Therefore, the difference between the magnitude of the displacement of the boy and the distance travelled by him which is OC is 10 kilometres.
Problem 7
Calculate the total displacement covered by a train before coming to rest, if it’s initial velocity is 30m/s with a constant retardation of 0.1ms-2.
Solution
Data
This is what i will do to help you understand how to find displacement in physics most especially from the above question.
I will extract my data from the question first, and then then apply the right formula to solve the problem.
Initial velocity, u = 30m/s
Retardation is the negative of acceleration, and this implies that a = -0.1ms-2
As the train comes to rest, its final velocity, v = 0
By applying one of the equations of motion, we have
v2 = u2 + 2as
Which is equal to
0 = u2 + 2as
By making s subject of the formula, we have
2as = -u2
Which can also be written as
s = -u2/2a
Therefore, if we substitute our data into the above formula, we will end up with
s = – [ 302/ (2x -0.1) ]
negative will cancel negative on the RHS, and we will have
s = 900 / 0.2 = 4,500m
Therefore, the total displacement of the train before coming to rest is 4,500 meters.
How to Find Displacement in Physics: Real-World Examples
Example 1: Displacement in Everyday Life
Imagine you’re taking a stroll in the park. You start at the entrance (A) and end up near a beautiful fountain (B). The straight-line distance between A and B is the displacement, considering direction as well.
Example 2: Displacement in Travel
Let’s say you embark on a road trip. You start in your hometown (P) and drive to a picturesque hill station (Q). The displacement between P and Q is the direct path between the two, regardless of the actual route you took.
Applications of Displacement
The concept of displacement finds applications in various fields, including physics, engineering, and navigation.
Physics and Kinematics
In physics, displacement is a fundamental concept in kinematics—the study of motion. It helps us analyze an object’s change in position, velocity, and acceleration over time.
Engineering and Design
Engineers use displacement calculations to design efficient machinery and structures. By understanding how objects move and interact, engineers can create safer and more effective designs.
Navigation and GPS
Global Positioning System (GPS) devices rely on displacement calculations to determine your precise location. By tracking your movement and changes in position, GPS devices provide accurate navigation.
Video 2: Acceleration, Velocity, and Displacement
How to Find Displacement in Physics: FAQs
Q: What’s the difference between displacement and distance?
A: Distance is the total length of the path travelled by an object, while displacement considers both distance and direction between the initial and final positions.
Q: Can displacement be negative?
A: Yes, displacement can be negative when an object moves in the opposite direction of the positive reference point.
Q: Is displacement always shorter than distance?
A: Not necessarily. Displacement can be shorter, equal to, or longer than the distance travelled, depending on the object’s path.
Q: How is displacement represented graphically?
A: Displacement can be represented by an arrow on a coordinate plane, where the length of the arrow represents the magnitude, and the direction indicates the movement.
Q: Is displacement scalar or vector?
A: Displacement is a vector quantity because it has both magnitude and direction.
Q: What’s the SI unit of displacement?
A: The SI unit of displacement is the meter (m).
Therefore, I explained this topic in detail to make you understand how to find displacement in physics.
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