Contents
Distance & Displacement – Definitions, Differences and Worksheets
By going through this article, you will get a thorough knowledge about the distance as well as the displacement, and how to find it from the graphs, with the help of practice problems.
Distance and Displacement – Definitions
The total path covered by a particle in a given interval of time is called the distance travelled by that particle.
Distance is a scalar quantity. For a moving body, the distance covered is always positive. The SI unit for distance is metre and its dimensional formula is [M^{0}L^{1}T^{0}].
Consider a particle moving from P to R via Q as shown in figure 1.
Then, the distance covered = PQ + QR
Displacement may be defined as the change in position of a particle along a given direction.
The displacement is the shortest distance between the two points. It is obtained by drawing a straight line vector from initial to final positions. In figure 1, is the displacement. It is a vector quanity and is represented by . It can have any value i.e., a zero, –ve or +ve. If the particle is travelling from R to P along any path, the displacement is represented . Then .
If the particle moves from P to R and then from R to P along any path, the net displacement .
The unit and dimensions of displacement are the same as that for the distance which is metre and [ M^{0}L^{1}T^{0} ] respectively.
Examples:

In the case of a body circulating around a circle of radius r, when it completes a semicircle AB, the distance covered is πr, while displacement is 2r. When it completes one revolution, the distance covered is 2πr and the displacement is zero.
 Consider an object projected vertically from the top of a building of height h. The object moves a distance l upwards and returns to the ground. Now the distance travelled is (2l + h) while the displacement is –h.
Features of displacement
 Displacement is a vector quantity. It can be a positive,a negative or a zero value.
 The magnitude of displacement of a particle between two points gives the shortest distance between the two points.
 The SI unit for displacement is metre and it has the dimension of length [M^{0}L^{1}T^{0}]
 The magnitude of the displacement will be less than or equal to the actual distance travelled by the object in the given interval of time.
i.e., Displacement ≤ Distance  If an object, after travelling a certain distance returns to the starting point, then its displacement is zero.
 The displacement of the object between two points has a unique value.
 The displacement of an object is unaltered due to a shift in the origin of the position axis.
 The numerical ratio of displacement to distance is equal to or less than one.
Difference between distance and displacement in tabular form
Sl. No  Distance  Displacement 
1.  The distance travelled is the total path covered by a particle in a given interval of time. i.e., Δd = d_{1} + d_{2} + d_{3} + ……..+ d_{n} , where Δd is the total distance, d_{1}, d_{2}, d_{3}…..d_{n} are the paths covered by the particle to reach from initial to the final position. 
Displacement is the shortest distance from the initial to the final position of the body while considering the direction. i.e., Δ𝑥 = 𝑥_{2} – 𝑥_{1}_{ }, where Δ𝑥 is the displacement, 𝑥_{1 }and𝑥_{2}_{ }are the initial and final positions respectively. 
2.  It is denoted as ‘ d ’  It is denoted as ‘ s ’ 
3.  It is a scalar quantity. That is, the direction is not considered while calculating distance.  It is a vector quantity. That is, the direction is taken into consideration while calculating displacement. 
4.  It depends only upon the magnitude.  It depends upon both magnitude and direction. 
5.  The formula for distance is Speed × Time  The formula for displacement is Velocity × Time 
6.  The distance travelled by a body is always positive and can never be negative.  The displacement can have any value i.e., positive, negative or zero. 
7.  The distance depends upon the path travelled by the body and will vary according to the change in path.  The displacement does not depend on the path, while it depends only upon the initial and final positions. 
8.  The distance travelled is either equal to or greater than displacement and is never less than the magnitude of displacement. i.e., Distance ≥ Displacement 
The magnitude of displacement is less than or equal to the distance travelled during the course of motion. i.e., Displacement ≤ Distance 
9.  The distance can never decrease as time increases.  The displacement may increase or decrease as time increases. 
10.  The distance can be measured along a curved or a nonstraight line path.  The displacement is always measured along a straightline path. 
11.  If an object, after travelling a certain distance returns to the starting point, then the distance travelled will be the total path covered by the object and can never be zero.  If an object, after travelling a certain distance returns to the starting point, then its displacement is zero. 
12.  The distance between two points can have different values.  The displacement of the body between two points has a unique value. 
13.  The distance is never indicated with an arrow.  The displacement is always indicated with an arrow vector. 
14.  SI Unit and dimension for distance is metre and [ M^{0}L^{1}T^{0} ] respectively.  SI Unit and dimension for displacement is the same as that of distance which is metre and [M^{0}L^{1}T^{0}] respectively. 
Distance and Displacement problems
Problem 1:
A boat sailing through a river moved eastward for 5 km, then cross the river by moving 3 km southward. On reaching the other side it moved westward through 1 km and reached the jetty. Find the distance covered and displacement of the boat.
Solution:
From the fig, AB = 5 km
ED = BC = 3 km
CD = 1 km
∴ AE = 5 km – 1 km = 4 km
Distance covered = AB + BC + CD
= 5km + 3km + 1 km
= 9 km
So the distance covered by the boat is 9 km and the displacement is 5 km
Problem 2:
A car moving along in a straight highway from point P to point Q to point R and to point S, then back to point Q and finally to the point R as shown in the figure below.
a) Find the distance travelled by car.
b) Find the displacement of the car.
Ans:
Given the distances, PQ = 3 km, QR = 5 km and RS = 7 km
Also, SQ = 7 + 5 =12 km,
PR = 3 + 5 = 8 km
a) Distance travelled by the car = PQ + QR + RS + SQ + QR
= 3 + 5 + 7 + 12 + 5
= 32 km
b) Displacement of the car = the shortest distance between the final point R and the initial point P
= PR
= 8 km
Problem 3:
A person walks along the path of a rectangle from point P to point R as shown in the below figure.
a) Find the distance travelled by the person.
b) Find out the magnitude of the displacement of the person.
Ans:
Given the distances, PQ = 5 km,
QR = 2 km
a) The distance travelled by the person = PQ + QR
= 5 + 2
= 7 km
b) The magnitude of displacement is equal to the shortest distance between the final point R and the initial point P, which is equal to the diagonal PR and can be calculated using Pythagora’s theorem.
i.e., magnitude of displacement = PRBy Pythagora’s theorem, PR^{2} = PQ^{2} + QR^{2}
Problem 4:
A motorcycle rides from point P to Q to R to S and finally to P in a circular path as shown in the below figure.
Find a) the distance travelled by motorcycle.
b) the displacement.
Answer:
Given radius, r = 5 km
a) Here, the distance travelled by motor cycle = circumference of the circle (∵ the motor cycle moves one complete rotation)
= 2πr
= 2 × 3.14 × 5
= 31.4 kmb) Here, the initial point is P and the final point is also P. Therefore, there will be no change in position and hence the displacement is equal to zero.
Problem 5:
A vehicle moves from point P to Q to R to S in a circular path as shown in the below figure.
a) Find the distance travelled by the vehicle.
b) Find out the magnitude of the displacement of the vehicle.
Ans:
Given, radius, r = 8 km
a) Here, the vehicle moves only 3/4 th of one rotation.
∴, the distance travelled by the vehicle = 3/4 th of the circumference of the circle.
= (3/4)2πr
= (3/4) × 2 × 3.14 × 8
= 37.68 kmb) Here the initial point is P and the final point is S.
∴, The magnitude of displacement is equal to the shortest distance between the final point S and the initial point P, which is equal to the distance PS and can be calculated using Pythagoras theorem to the triangle POS as shown in the below figure.
Problem 6:
Consider an object moving in a straight line. The distances travelled by the object from the origin with respect to time are shown in the figure given below. The different parts of its motion are represented by P, Q, R, S, T, U, V and W. Find a) The distance travelled by the object in the first 2 seconds. b) the distance travelled by the object in the first 4 seconds. c) the distance travelled by the object in the first 6 seconds. d) the distance travelled by the object in the first 8 seconds. e) the distance travelled by the object in the first 9 seconds. f) total distance travelled by the object in 14 seconds and g) displacement in 14 seconds.
Ans:
From the distance time graph provided in the question, we can find out the distance and displacement for different time intervals. From the fig,
a) The distance travelled by the object in first 2 seconds = 60 m
b) The distance travelled by the object in first 4 seconds = 90 m
c) The distance travelled by the object in first 6 seconds = 90 m
d) The distance travelled by the object in first 8 seconds = 150 m
e) Distance travelled by the object in 9 seconds = 150 + (150 – 120 )
= 150 m + 30 m
= 180m.
That is, from t = 0 to t = 8 seconds, the object has moved a distance of 150 m from the origin and from t = 8 to t = 9 sec, the object has travelled back a distance of 30 meters. So the total distance will be 150 + 30 = 180 m.
f) The distance travelled by the object in 14 seconds = Total distance travelled by the object.
= 150 + 30 + 120
= 300 m.
That is, from t = 0 to t = 8 seconds, the object has moved a distance of 120 m from the origin and from t = 8 to t = 14 sec, the object has come back to its initial position and travelled a distance of 150 meters again. So total distance travelled = 150 m + 150 m = 300 metre.g) Since the object has come back to its initial position, the total displacement is zero.
Problem 7:
The movement of objects P, Q, R, S and T is marked on a scale as shown in the below figure. Find the distance and displacement covered by each object.
Answer:
a) Consider the object P
Object P had an initial position of 1 metre and a final position of 4 metres.
∴ displacement of object P, Δx_{p }= final position – initial position
= 4 – 1
= +3 metres.
Distance travelled by object P = total path covered covered by P
= 3 metresb)Consider the object Q
Object Q had an initial position of 11 metres and a final position of 7 metres.
∴ displacement of object Q, Δx_{q }= final position – initial position
= 7 – 11
= 4 metres.
Distance travelled by object Q = total path covered by Q
= 4 metresc) Consider the object R
Object R had an initial position of 0 metres and a final position of 6 metres.
∴ displacement of object R, Δx_{r}= final position – initial position
= 6 – 0
= +6 metres.
Distance travelled by object R = total path covered by R
= 6 + 3 + 3
= 12 metresd) Consider the object S
Object S had an initial position of 8 metre and final position of 7 metres.
∴ displacement of object S, Δx_{s}= final position – initial position
= 7 – 8
= –1 metre.
Distance travelled by object S = total path covered by S
= 3 + 4
= 7 metrese) Consider the object T
Object T had an initial position of 7 metres and final position of 8 metres.
∴ displacement of object T, Δx_{t }= final position – initial position
= 8 – 7
= 1 metre.
Distance travelled by object T = total path covered by T
= 4 + 3
= 7 metres.
Problem 8:
The horizontal position of a car in kilometres over time is shown below.
a) Find the displacement and the distance travelled by car between 1 hour and 3 hours.
b) What is the displacement and distance covered by the car between 3 hours and 5 hours?
c) Calculate the displacement and the distance travelled by car between 5 hours and 9 hours.
d) Find the displacement and the distance covered by the car between 3 hours and 9 hours.
e) Find the total displacement and the distance covered by the car.
Answer:
From the positiontime graph given, we can calculate the distance and displacement for different time intervals.
a) Between 1 hour an 3 hour, the car had an initial position of 40 km and a final position of 40 km.
∴Displacement of the car between 1 hour and 3 hours,Δx = final position – initial position = 40 – 40 =0 km.Distance travelled by car between 1 hour and 3 hours = total path covered between 1 hour and 3 hours =0 km( Since that portion of the graph is a straight line parallel to the xaxis).
b) Similarly, the car had an initial position of 40 km and a final position of 160 km between 3 hours and 5 hours.
∴ Displacement of the car between 3 hour and 5 hours, Δx = final position – initial position = 160 – 40 =120 km.Distance covered by the car between 3 hour and 5 hours= total path covered between 3 hours and 5 hours =120 km.
c) Similarly, the car starts at an initial position of 160 km and ends at a final position of 0 km during the period 5 hours and 9 hours.
∴ Displacement of the car between 5 hours and 9 hours, Δx = final position – initial position = 0 – 160 = 160 km ( displacement is negative means the car moves in the opposite or negative direction).Distance covered by the car between 5 hours and 9 hours = total path covered between 5 hours and 9 hours = 160 km (distance is always positive)
d) Similarly, between the time 3 hours and 9 hours, the initial and final position of the car is 40 km and 0 km respectively.
∴ Displacementof the car between 3 hours and 9 hours,Δx = final position – initial position = 0 – 40 = 40 km ( displacement is negative means the car moves in the opposite or negative direction).From the graph, it is clear that the car travels in two segments between 3 hours and 9 hours. i.e., the car starts at 40 km at 3 hours and moves to 160 km at 5 hours, travelling a distance of 120 km.Also, the car starts at 120 km at 5 hours and moves to 0 km at 9 hours, travelling a distance of 160 km.
∴ Distancecovered by the car between 3 hours and 9 hours = total path covered between 5 hours and 9 hours = 120 km + 160 km = 280 km.e) While considering the total motion, the car starts and ends at the same position of 0 km. That is, its initial position is 0 km at 0 hour and final position is also 0 km at 9 hours.
∴ Total displacement of the car, Δx = Displacement of the car between 0 hours and 9 hours= final position – initial position = 0 – 0 = 0 km.Also, while considering the total motion, it is clear that the car travels in four segments between 0 hours and 9 hours. i.e., the car starts at 0 km at 0 hours and moves to 40 km at 1 hour, travelling a distance of 40 km. Also, the car is stationary between the time 1 hour and 3 hours, thus the distance covered in this segment is 0 km. Again, the car starts at 40 km at 3 hours and moves to 160 km at 5 hours, travelling a distance of 120 km. Finally, the car starts at 120 km at 5 hours and moves to 0 km at 9 hours, travelling a distance of 160 km.
∴ Total distance covered by the car = total path covered between 0 hours and 9 hours = 40 + 0 + 120 + 160 = 320 km
Problem 9:
A boy rides a bicycle back and forth along the ground and the horizontal position of the bicycle is given below.
a) Find the displacement and the distance covered by the bicycle between 0 second and 30 seconds.
b) Calculate the displacement and distance travelled by bicycle between 0 second and 40 seconds
c) Calculate the displacement and distance travelled by bicycle between 30 seconds and 50 seconds.
d) What is the total displacement and distance travelled by bicycle?
Answer:
The distance and displacement for different time intervals can be found out from the position vs time graph given.
a) The bicycle had an initial position of 60 metres and a final position of –30 metres between 0s and 30 seconds.
∴ Displacement of the bicycle between 0 second and 30 seconds, Δx = final position – initial position = 30 – 60 = 90 m.Also, from the pt graph given, it is clear that the bicycle travels in two segments between 0 second and 30 seconds. i.e., the bicycle starts at 60 m at 0 second and moves to – 30 metres at 15 seconds, travelling a distance of 90 m. Also, the bicycle is stationary between the time 15 seconds and 30 seconds, thus the distance travelled in this segment is 0 m.
∴ Distance covered by the bicycle between 0 second and 30 seconds = total path covered between 0 seconds and 30 seconds = 90 + 0 = 90 m.b) Similarly, between the time 0 seconds and 40 seconds, the bicycle had an initial and final position of 60 metres and 30 metres respectively.
∴ Displacement of the bicycle between 0 second and 40 seconds, Δx = final position – initial position = 30 – 60 = 30 m.Also, from the graph, it is clear that the bicycle travels in three segments between 0 second and 40 seconds. i.e., the bicycle starts at 60 m at 0 second and moves to – 30 metres at 15 seconds, travelling a distance of 90 m. Also, the bicycle is stationary between the time 15 seconds and 30 seconds, thus the distance travelled in this segment is 0 m. Again, the bicycle starts at –30 metres at 30 seconds and moves to 30 metres at 40 seconds, covering a distance of 60 metres.
∴ Distance covered by the bicycle between 0 second and 40 seconds = total path covered between 0 seconds and 40 seconds = 90 + 0 + 60 = 150 m.c) Similarly, the bicycle starts at an initial position of 30 metres and ends at a final position of 0 metres during the period 30 seconds and 50 seconds.
∴ Displacement of the bicycle between 30 second and 50 seconds, Δx = final position – initial position = 0 – 30 = +30 m.Also, from the graph, it is clear that the bicycle travels in two segments between 30 second and 50 seconds. i.e., the bicycle starts at 30 m at 30 seconds and moves to 30 metres position at 40 seconds, travelling a distance of 60 m. Also, the bicycle starts at 30 metres at 40 seconds and moves to 0 metres position at 50 seconds, covering a distance of 30 metres.
∴ Distance covered by the bicycle between 0 second and 50 seconds = total path covered between 0 seconds and 50 seconds = 60 + 30 = 90 m.d) Now while considering the total motion, the bicycle starts at 60 metres and ends at 0 metres.
∴ Total displacement of the car, Δx = Displacement of the car between 0 second and 50 seconds = final position – initial position = 0 – 60 = 60 m.Also, from the graph, it is clear that the bicycle travels in four segments between 0 second and 50 seconds. i.e., the bicycle starts at 60 m at 0 second and moves to – 30 metres at 15 seconds, travelling a distance of 90 m. Also, the bicycle is stationary between the time 15 seconds and 30 seconds, thus the distance travelled in this segment is 0 m. Again, the bicycle starts at –30 metres at 30 seconds and moves to 30 metres at 40 seconds, covering a distance of 60 metres. Finally, the bicycle starts at 30 metres at 40 seconds and moves to a 0 metre position at 50 seconds, covering a distance of 30 metres.
∴ Total distance covered by the bicycle= total path covered between 0 second and 50 seconds = 90 + 0 + 60 + 30 = 180 m.
Problem 10:
The positiontime graph for an elevator travels up and down is given below. Find the distance and displacement of the elevator between 6 seconds and 21 seconds.
Answer:
The elevator had an initial position of 15 metres and a final position of 20 metres between 6 s and 21 seconds.
∴ Displacement of the elevator between 6 seconds and 21 seconds, Δx = final position – initial position = 20 – 15 = 35 m.Also, from the graph, it is clear that the elevator moves in two segments between 6 second and 21 seconds. i.e., the elevator starts at 15 m at 6 second and moves to a 0metre position at 9 seconds, covering a distance of 15 m. Again, the elevator starts at 0 m at 9 seconds and moves to 20 metres position at 21 seconds, covering a distance of 20 m.
∴ Distance covered by the elevator between 6 seconds and 21 seconds= total path covered between 6 seconds and 21 seconds = 15 + 20 = 35 m.
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Also, learn the difference between speed and velocity.