This tutorial discusses important conceptual examples of distance. And the answer is given with the example. So, you will have no difficulty in understanding.
1. If displacement of a particle is zero, then distance will be
If the displacement of a particle is zero, then distance may or may not be zero.
In both cases, the displacement of any particle will be zero.
1st case: When a particle is stationary, both displacement and distance will be zero.
2nd case: A particle will keep moving. But, initial point and final point are the same. In that case, the displacement of the particle is zero but the distance is not zero.
2. If the distance covered is zero, then displacement will be
Distance zero means the particle is stationary. That is, in this case, the displacement of a particle must be zero.
3. If a particle covers a minimum distance, then the relationship between distance(d) and displacement(s) will be
The minimum distance between the two points will be equal to the linear distance between the two points.
By definition, the linear distance between two points is called displacement. Thus, in this case, distance and displacement will be equal.
4. If a particle moves in a curved path. In that case, what will be the relation between distance(d) and displacement(s) of a particle?
When a particle moves along a curved path, the distance will always be greater than the displacement. Because displacement is the linear distance between two points.
5. Suppose, you move from point a to point b along a semicircle path. Then, what will be your displacement and distance?
If a particle moves along a semicircular path from point A to point B, the distance traveled by the particle will be πr.
If the distance of the particle is πr then the displacement of the particle will be equal to the diameter of the semicircle.
6. The ratio of displacement and distance traveled by an object can never be
The displacement will be equal to or smaller than the distance, but not larger. For this reason the ratio of displacement and distance will not be greater than 1.
7. A runner completes one round of a circular path of radius r in 40 seconds. His displacement and distance after 2 minutes 20 seconds will be
If the runner completes one round in 40 seconds, he will complete two and a half rounds in 2 minutes and 20 seconds.
You know, moving from the initial point back to the initial point, the displacement will always be zero.
Thus, the runner’s displacement will be zero for the complete two rounds. However, for the rest of the half round, the runner’s displacement will be equal to the diameter of the circular path.
s = 2r
And the runner’s total distance cover will be
d= (2+1/2)×2pr = 5pr
8. The equation of motion of an object in two-dimensional space is, x = 5t² + 2; y = 2t² + 5. The motion path of the object is
According to the question, x = 5t² + 2, y = 2t² + 5.
You put the value of t² in the equation of y.
So, you look at the equation above, it’s a straight line equation. Then this moving particle will move along a straight line.
9. A body covered a distance of L m along a curved path of a quarter circle. The ratio of distance to displacement is
As shown in the figure above, suppose the radius of the quarter circle is r.
Then the total distance will be
And the displacement will be
Therefore, the ratio of displacement to distance is π/2√2.
10. A car running at a speed of 18 km/h, the distance covered in 3s is
The above question gives the value of speed and time. However, the units of speed and time are defined in two different ways.
Thus, the unit of speed has to be converted in the same method according to the unit of time and answer. Here the unit of time and answer is given in the SI method.
v = 18 ×1000 /3600 = 5 m.s-1
Therefore, in three seconds the distance of the car will be
d = vt = 15 m
11. Two bodies of different masses ma and mb are dropped from two different heights, viz a and b. The ratio of times taken by the two to drop through these distances is
Notice the equation above, time does not depend on the mass of the object. So, for a and b heights, the total time taken is
Therefore, the ratio will be two different times
12. The equation of velocity of an object v = at. The distance that the particle travels in the first 6s is
According to the question, if v = at, then the total distance between the first six seconds will be
13. An object gains v velocity after crossing a distance h when it falls freely from a rest on the gravitational surface. And how far will the velocity be 2v if the object travels?
v² = 2gh
Assuming that the object gains 2v speed then its distance is h’.
(2v)² = 2gh’
h’ = 4h [ v² = 2gh ]
14. A body starting from rest and has uniform acceleration 8 m/sec2. The distance traveled by it in 6th second will be
Sn = u + ½(2n-1)
Then according to the question, in the 6th second the distance of the object will be
S6 = ½ × 8 × 11 [ u =0]
∴ S6 = 44
15. A body starts from rest, the ratio of distances traveled by the body during the 5th and 6th seconds is
S5 = ½ × a × 9
S6 = ½ × a × 11
Thus, the ratio between the two distances will be S5 : S6 = 9 : 11
16. The distance traveled by a particle starting from rest and moving with an acceleration 4/3 m/s2 in the third second is
S3 = ½ × 4/3 × (6-1) [ u =0]
∴ S3 = 10/3m
17. The initial velocity of a particle is 10 m/sec and its retardation is 2 m/sec2. The distance covered in the fifth second of the motion will be
Sn = u + ½a(2n-1)
According to the question, u =10 m/s and a = – 2 m/s², than
S5 = 10 – ½×2×9
∴ S5 = 1
18. A particle moves with constant acceleration for 6 seconds after starting from rest. The distance traveled during the consecutive 2 seconds interval are in the ratio
Sn = 2u + 2a(n-1)
Then, the distance covered in 2nd, 4th and 6th seconds will be
S2 = 2a(n-1) = 2a(2-1) = 2a
S4 = 2a(n-1) = 2a(4-1) = 6a
S6 = 2a(n-1) = 2a(6-1) = 10a
∴ S2 : S4 : S6 = 1 : 3 : 5
19. A particle moves on a straight line and velocity varies with time as v = 4t, find the distance traveled by the particle in the interval of 2 s to 4s
20. If a particle travels from rest at uniform acceleration and crosses the x path in first 2 s and the y path in next 2 s, then which of the following is correct?
If the particle has moved from a rest. Then we can write that,
21. A rock is falling freely under the influence of gravity. h1, h2, and h3 distances cross at intervals of the first, second and third 5s of the fall period. Then the relationship between h1, h2 and h3 will be
At a five second interval, the total distance in nth second will be
22. An object starts its journey from rest at constant acceleration and crosses the x distance in first 10s and the y distance in next 20s. The relationship between x and y is
23. The distance traveled by a moving particle is equal to half of the product of instantaneous velocity and time. Which of the following statements is correct?
24. An object is released from some height. Exactly after one second, another object is released from the same height. The distance between the two objects exactly after 2 seconds of the release of second object will be
h1=1/2 × g × 3²
h2=1/2 × g × 2²
Then the distance between two balls will be
∴ h1- h2= 1/2 × g × 5 = 24.5 m
25. Two balls are dropped to the ground from different heights. One ball is dropped 2 sec after the other but they both strike the ground at the same time. If the first ball takes 3 sec to reach the ground then the difference in initial heights is:
This question is similar to the question above, the second ball started journey two seconds after the first ball. And if the first ball travels five seconds, then the second ball travels three seconds.
Then the distance between two balls will be
26. A particle moves along a straight line OX. At a time t (in seconds ) the distance x (in meters) of the particle from O is given by
x = 40 + 12t – t3
How long would the particle travel before coming to rest?
Once you think of a moving object, when it will come to a rest! Yes, when the velocity of a moving object becomes zero, the moving object will come to rest. So, the equation of time with distance in question is given.
x = 40+12t -t³
When the object is at rest, the derivative will be zero in respect of time.
Therefore, distance will be
x = 40 + 24 -8 = 56m
27.A body moves along the curved path of a quarter circle. Calculate the ratio of distance to displacement
28. The variation of velocity of a particle moving along a straight line is shown in the figure. The distance traversed by the body in 4 seconds is
0-1s: (1/2 × 1 × 20) = 10
1-2s: (1 × 20) = 20
2-3s: (1/2 × 1 × 10) + (1 × 10) = 15
3-4s: (1 × 10) = 10
Total distance cover in 0-4 seconds is 55 m.
29. The velocity-time graph of a linear motion is shown below. The displacement from the origin after 8 seconds is
30. If a particle is moving at a speed of v=|t-t0| for 2t0 seconds, what is the distance traveled by the particle?
The area of v-t according to the graph in first 2t0 seconds will be
d = 1/2 × t0 × t0 + 1/2 × t0 × t0 = t²0
31. A bird is flying in the sky along a straight line at v=|t-3| m/s speed. After 6 seconds the distance covered by the bird will be
If there is any mistake between the above question and answer, please let us know in the comment box below. The mistake will be corrected.