What is Displacement in Physics
In this post, i will help you to understand how to find displacement in physics.
Definition of Displacement in Physics: Displacement is the measure of seperation between two points in a specified direction. Displacement is a vector quantity and is measured in meters.
When a person moves from one point to another (changes his position) in straight, specified direction. We measure his distance as displacement. Displacement is direct, staright, and in 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-kilometers. After reaching town Z, the train followed the same route and stopped at 80-kilometers 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 measure of separation between two points. However, displacement is the exact opposite of distance, this is because displacement is in specified direction while distance is not.
Displacement is related to velocity while distance is related to speed. They are both related to time.
Formula for Calculating Displacement in Physics
The formula for displacement can be:
- 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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Examples of How to Find Displacement in Physics
Here are some examples to help you understand how to find displacement in physics.
Example 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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Example 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
Example 3
A body starts with an initial velocity of 4 m/s and continous 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 dont 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 dispalcement by applying the formula
Dispacement, S = Average Velocity x Time
Thus,
Displacement = 8.5 x 6 = 51m
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Example 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.
Example 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
Distance covered by the car is the displacement of the bus. This 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.
Example 6
A boy travels 8-kilometer eastward to a point B and then 6-kilometer 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 distance travelled by him which is OC, is 10-kilometers.
Example 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, it’s 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.
Therefore, I explained this topic in detail to make you understand how to find displacement in physics.
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