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Course by: Sunil Kumar Singh. E-mail the author

# Graphs of motion with constant acceleration

Module by: Sunil Kumar Singh. E-mail the author

Graphical analysis of motion with constant acceleration in one dimension is based on the equation defining position and displacement, which is a quadratic function. Nature of motion under constant acceleration and hence its graph is characterized by the relative orientation of initial velocity and acceleration. The relative orientation of these parameters controls the nature of motion under constant acceleration. If initial velocity and acceleration are in the same direction, then particle is accelerated such that speed of the particle keeps increasing with time. However, if velocity and acceleration are directed opposite to each other, then particle comes to rest momentarily. As a result, the motion is divided in two segments - one with deceleration in which particle moves with decreasing speed and second with acceleration in which particle moves with increasing speed.

We study various scenarios of motion of constant acceleration by analyzing equation of position, which is quadratic expression in time. The position of particle is given by :

x = x 0 + u t + 1 2 a t 2 x = x 0 + u t + 1 2 a t 2 x = 1 2 a t 2 + u t + x 0 x = 1 2 a t 2 + u t + x 0

Clearly, the equation for position of the particle is a quadratic expression in time, “t”. A diagram showing a generalized depiction of motion represented by quadratic expression is shown here :

In case initial position of the particle coincides with origin of reference, then initial position x 0 = 0 x 0 = 0 and in that case x denotes position as well as displacement :

x = u t + 1 2 a t 2 x = u t + 1 2 a t 2

## Nature of graphs

The nature of quadratic polynomial is determined by two controlling factors (i) nature of coefficient of squared term, t 2 t 2 and (ii) nature of discriminant of the quadratic equation, which is formed by equating quadratic expression to zero.

### Nature of coefficient of squared term

The coefficient of squared term is “a/2”. Thus, its nature is completely described by the nature of “a” i.e. acceleration. If “a” is positive, then graph is a parabola opening up. On the other hand, if “a” is negative, then graph is a parabola opening down. These two possibilities are shown in the picture.

It may be interesting to know that sign of acceleration is actually a matter of choice. A positive acceleration, for example, is negative acceleration if we reverse the reference direction. Does it mean that mere selection of reference direction will change the nature of motion? As expected, it is not so. The parabola comprises two symmetric sections about a line passing through the minimum or maximum point (C as shown in the figure). One section represents a motion in which speed of the particle is decreasing as velocity and acceleration are in opposite directions and second section represents a motion in which speed of the particle is increasing as velocity and acceleration are in the same direction. In other words, one section represents deceleration, whereas other section represents acceleration. The two sections are simply exchanged in two graphs. As such, a particular motion of acceleration is described by different sections of two graphs when we change the sign of acceleration. That is all. The nature of motion remains same. Only the section describing motion is exchanged.

### Nature of discriminant

Comparing with general quadratic equation A x 2 + B x + C = 0 A x 2 + B x + C = 0 , we see that determinant of the corresponding quadratic equation is given by :

D = B 2 4 A C = u 2 4 X a 2 X x 0 D = B 2 4 A C = u 2 4 X a 2 X x 0 D = u 2 2 a x 0 D = u 2 2 a x 0

The important aspect of discriminant is that it comprises of three variable parameters. However, simplifying aspect of the discriminant is that all parameters are rendered constant by the “initial” setting of motion. Initial position, initial velocity and acceleration are all set up by the initial conditions of motion.

The points on the graph intersecting t-axis gives the time instants when particle is at the origin i.e. x=0. The curve of the graph intersects t-axis when corresponding quadratic equation (quadratic expression equated to zero) has real roots. For this, discriminant of the corresponding quadratic equation is non-negative (either zero or positive). It means :

D = u 2 2 a x 0 0 D = u 2 2 a x 0 0 2 a x 0 u 2 2 a x 0 u 2

Note that squared term u 2 u 2 is always positive irrespective of sign of initial velocity. Thus, this condition is always fulfilled if the signs of acceleration and initial position are opposite. However, if two parameters have same sign, then above inequality should be satisfied for the curve to intersect t-axis. In earlier graphs, we have seen that parabola intersects t-axis at two points corresponding two real roots of corresponding quadratic equation. However, if discriminant is negative, then parabola does not intersect t-axis. Such possibilities are shown in the figure :

Clearly, motion of particle is limited by the minimum or maximum positions. It is given by :

x = D 4 A = u 2 2 a x 0 4 X a 2 = u 2 2 a x 0 2 a x = D 4 A = u 2 2 a x 0 4 X a 2 = u 2 2 a x 0 2 a

It may be noted that motion of particle may be restricted to some other reference points as well. Depending on combination of initial velocity and acceleration, a particle may not reach a particular point.

Our consideration, here, considers only positive values of time. It means that we are discussing motion since the start of observation at t=0 and subsequently as the time passes by. Mathematically, the time parameter, “t” is a non-negative number. It can be either zero or positive, but not negative. In the following paragraphs, we describe critical segments or points of a typical position – time graph as shown in the figure above :

Point A : This is initial position at t = 0. The position corresponding to this time is denoted as x 0 x 0 . This point is the beginning of graph, which lies on x-axis. The tangent to the curve at this point is the direction of initial velocity, “u”. Note that initial position and origin of reference may be different. Further, this point is revisited by the particle as it reaches E. Thus, A and E denotes the same start position - though represented separately on the graph.

Curve AC : The tangent to the curve part AB has negative slope. Velocity is directed in negative x-direction. The slope to the curve keeps decreasing in magnitude as we move from A to C. It means that speed of the particle keeps decreasing in this segment. In other words, particle is decelerated during motion in this segment.

Points B, D : The curve intersects t-axis. The particle is at the origin of the reference, O, chosen for the motion. The time corresponding to these points (B and D) are real roots of the quadratic equation, which is obtained by equating quadratic expression to zero.

Point C : The slope of tangent to the curve at this point is zero. It means that the speed of the particle has reduced to zero. The particle at this point is at rest. The slopes of the tangent to curve about this point changes sign. It means that velocity is oppositely directed about this point. The point B, therefore, is a point, where reversal of direction of motion occurs. Note that particle can reverse its direction only once during its motion under constant acceleration.

Point E : The particle reaches start point E (i.e. A) again in its motion after reversal of motion at C.

Point F : This is the end point of motion.

The graph of a motion under constant acceleration is bounded by set of parameters defining a motion. In a particular case, we may be considering only a segment of the curve starting from point A and ending at point F – not necessarily covering the whole of graph as shown in the figure. The point F may lie anywhere on the graph. Further, nature of curve will be determined by values and signs of various parameters like initial position, initial velocity and acceleration. Here, we have divided our study in two categories based on the sign of acceleration : (i) acceleration is positive and (ii) acceleration is negative.

## Acceleration is positive (in the reference direction)

The graph of quadratic equation is a parabola opening upwards as coefficient of squared term t 2 t 2 is positive i.e. a > 0. The minimum value of expression i.e. x is :

x min = D 4 A = u 2 2 a x 0 4 X a 2 = u 2 2 a x 0 2 a x min = D 4 A = u 2 2 a x 0 4 X a 2 = u 2 2 a x 0 2 a

The various possibilities of initial positions of the particle with time are discussed here :

Case 1: Initial position and origin of reference are different. Initial position is positive. Initial velocity is negative i.e. it is directed in negative reference direction. The velocity and acceleration are oppositely directed.

Initially particle is decelerated as the speed of the particle keeps on decreasing till it becomes zero at point C. This is indicated by the diminishing slope (magnitude) of the tangents to the curve. Subsequently, particle is accelerated so long force causing acceleration is applied on the particle.

The segment DF with origin at D is typical graph of free fall of particle under gravity, considering DF as the height of fall and downward direction as positive direction. Only differing aspect is that particle has initial velocity. Nevertheless, the nature of curve of free fall is similar. Note that speed of the particle keeps increasing till it hits the ground.

#### Example 1

Problem : Given x 0 = 10 m ; u = - 15 m / s ; a = 10 m / s 2 x 0 = 10 m ; u = - 15 m / s ; a = 10 m / s 2

Find the position when particle reverses its motion.

Solution : Acceleration is negative. The position-time graph is a parabola opening up. Therefore, the particle changes its direction of motion at its minimum position, which is given as :

x min = - u 2 2 a x 0 2 a x min = - u 2 2 a x 0 2 a

x min = - 15 2 2 X 10 X 10 2 X 10 = 25 20 = 1.25 m x min = - 15 2 2 X 10 X 10 2 X 10 = 25 20 = 1.25 m

Alternatively,

At the point of reversal, velocity is zero. Using v=u+at, :

0 = - 15 + 10 t 0 = - 15 + 10 t t = 3 2 = 1.5 s t = 3 2 = 1.5 s

Position of particle in 1.5 s is :

x = x 0 + u t + 1 2 a t 2 x = x 0 + u t + 1 2 a t 2 x = 10 + 15 X 1 .5 + 1 2 X 10 X 2.25 = 10 22.5 + 5 X 2.25 = 12.5 + 11.25 = 1.25 m x = 10 + 15 X 1 .5 + 1 2 X 10 X 2.25 = 10 22.5 + 5 X 2.25 = 12.5 + 11.25 = 1.25 m

#### Exercise 1

Given : x 0 = 10 m ; u = - 15 m / s ; a = 10 m / s 2 x 0 = 10 m ; u = - 15 m / s ; a = 10 m / s 2

Find the time instants when the particle is at origin of reference and initial position. Also find the time when velocity is zero.

##### Solution

The position of the particle with respect to origin of reference is given by :

x = x 0 + u t + 1 2 a t 2 x = x 0 + u t + 1 2 a t 2

In order to determine time instants when the particle is at origin of reference, we put x=0,

0 = 10 - 15 t + 1 2 X 10 t 2 0 = 10 - 15 t + 1 2 X 10 t 2 t 2 3 t + 2 = 0 t 2 3 t + 2 = 0 t 2 t 2 t + 2 = 0 t 2 t 2 t + 2 = 0 t t 1 2 t 1 = 0 t t 1 2 t 1 = 0 t = 1 s and 2 s t = 1 s and 2 s

In order to determine time instant when particle is at initial position, we put x=10,

10 = 10 - 15 t + 1 2 X 10 t 2 10 = 10 - 15 t + 1 2 X 10 t 2 5 t 2 - 15 t = 0 5 t 2 - 15 t = 0 t = 0 and 3 s t = 0 and 3 s

The particle is at the initial position twice, including start of motion. Now, at the point of reveral of direction, speed of the particle is zero. Putting v = 0 and using v=u+at,

0 = - 15 + 10 t 0 = - 15 + 10 t t = 1.5 s t = 1.5 s

Case 2: Initial position and origin of reference are different. Initial position is positive. Initial velocity is positive i.e. it is directed in reference direction. The velocity and acceleration are in same direction.

The particle is accelerated so long force causing acceleration is applied on the particle. The segement AF with origin at D is typical graph of free fall of particle under gravity, considering AF as the height of fall and downward direction as positive direction. Only differing aspect is that particle has initial velocity. Neverthless, the nature of curve of free fall is similar. Note that speed of the particle keeps increasing till it hits the ground.

#### Exercise 2

Given x 0 = 10 m ; u = 15 m / s ; a = 10 m / s 2 x 0 = 10 m ; u = 15 m / s ; a = 10 m / s 2

Find the time instants when the particle is at origin of reference and initial position. Also find the time when velocity is zero.

##### Solution

The position of the particle with respect to origin of reference is given by :

x = x 0 + u t + 1 2 a t 2 x = x 0 + u t + 1 2 a t 2

In order to determine time instants when the particle is at origin of reference, we put x=0,

0 = 10 + 15 t + 1 2 X 10 t 2 0 = 10 + 15 t + 1 2 X 10 t 2 t 2 + 3 t + 2 = 0 t 2 + 3 t + 2 = 0 t 2 + t + 2 t + 2 = 0 t 2 + t + 2 t + 2 = 0 t = - 1 s and - 2 s t = - 1 s and - 2 s

We neglect both these negative values and deduce that the particle never reaches point of origin. In order to determine time instant when particle is at initial position, we put x=10,

10 = 10 + 15 t + 1 2 X 10 t 2 10 = 10 + 15 t + 1 2 X 10 t 2 5 t 2 + 15 t = 0 5 t 2 + 15 t = 0 t = 0 and - 3 s t = 0 and - 3 s

We neglect negative value. The particle is at the initial position only at the start of motion. Now, at the point of reveral of direction, speed of the particle is zero. Putting v = 0 and using v=u+at,

0 = 15 + 10 t 0 = 15 + 10 t t = - 1.5 s t = - 1.5 s

The particle never changes its direction.

Case 3: Initial position and origin of reference are same. Initial velocity is negative i.e. it is directed in negative reference direction. The velocity and acceleration are oppositely directed.

Initially particle is decelerated as the speed of the particle keeps on decreasing till it becomes zero at point C. This is indicated by the diminishing slope (magnitude) of the tangents to the curve. Subsequently, particle is accelerated so long force causing accleration is applied on the particle.

#### Exercise 3

Given x 0 = 0 m ; u = 15 m / s ; a = 10 m / s 2 x 0 = 0 m ; u = 15 m / s ; a = 10 m / s 2

Find the time instants when the particle is at origin of reference. Also find the time when velocity is zero.

##### Solution

The position of the particle with respect to origin of reference is given by :

x = u t + 1 2 a t 2 x = u t + 1 2 a t 2

In order to determine time instants when the particle is at origin of reference, we put x=0,

0 = - 15 t + 1 2 X 10 t 2 0 = - 15 t + 1 2 X 10 t 2 t 2 - 3 t = 0 t 2 - 3 t = 0 t = 0 and 3 s t = 0 and 3 s

The particle is at the origin of reference twice, including start of motion. Now, at the point of reversal of direction, speed of the particle is zero. Putting v = 0 and using v=u+at,

0 = - 15 + 10 t 0 = - 15 + 10 t t = 1.5 s t = 1.5 s

Case 4: Initial position and origin of reference are same. Initial velocity is positive i.e. it is directed in reference direction. The velocity and acceleration are in same direction.

The particle is accelerated so long force causing acceleration is applied on the particle.

The segement OF is typical graph of free fall of particle under gravity, considering OF as the height of fall and downward direction as positive direction. Only differing aspect is that particle has initial velocity. Neverthless, the nature of curve of free fall is similar. Note that speed of the particle keeps increasing till it hits the ground.

#### Exercise 4

Given x 0 = 0 m ; u = 15 m / s ; a = 10 m / s 2 x 0 = 0 m ; u = 15 m / s ; a = 10 m / s 2

Find the time instants when the particle is at origin of reference. Also find the time when velocity is zero.

##### Solution

The position of the particle with respect to origin of reference is given by :

x = u t + 1 2 a t 2 x = u t + 1 2 a t 2

In order to determine time instants when the particle is at origin of reference, we put x=0,

0 = 15 t + 1 2 X 10 t 2 0 = 15 t + 1 2 X 10 t 2 t 2 + 3 t = 0 t 2 + 3 t = 0 t = 0 and - 3 s t = 0 and - 3 s

Neglecting negative value of time,

t = 0 t = 0

The particle is at the origin of reference only at the start of motion. Now, at the point of reversal of direction, speed of the particle is zero. Putting v = 0 and using v=u+at,

0 = 15 + 10 t 0 = 15 + 10 t t = - 1.5 s t = - 1.5 s

We neglect negative value and deduce that particle never ceases to move and as such there is no reversal of motion.

## Acceleration is negative (opposite to the reference direction)

In this case, the graph of quadratic equation is a parabola opening downwards as coefficient of squared term t 2 t 2 is negative i.e. a > 0. The maximum value of expression i.e. x is :

x max = D 4 A = u 2 2 a x 0 4 X a 2 = u 2 2 a x 0 2 a x max = D 4 A = u 2 2 a x 0 4 X a 2 = u 2 2 a x 0 2 a

The analysis for this case is similar to the first case. We shall, therefore, not describe this case here.

### Exercise 5

Given x 0 = 10 m ; u = 15 m / s ; a = 10 m / s 2 x 0 = 10 m ; u = 15 m / s ; a = 10 m / s 2

Find the time instants when the particle is at origin of reference and initial position. Also find the time when velocity is zero.

#### Solution

The position of the particle with respect to origin of reference is given by :

x = x 0 + u t + 1 2 a t 2 x = x 0 + u t + 1 2 a t 2

In order to determine time instants when the particle is at origin of reference, we put x=0,

0 = 10 + 15 t + 1 2 X - 10 X t 2 0 = 10 + 15 t + 1 2 X - 10 X t 2 - t 2 + 3 t + 2 = 0 - t 2 + 3 t + 2 = 0

t = - 3 ± { 9 4 X 1 X 2 } 2 X 1 t = - 3 ± { 9 4 X 1 X 2 } 2 X 1

t = - 3 ± 4.12 2 = - 0.56 s or 3.56 s t = - 3 ± 4.12 2 = - 0.56 s or 3.56 s

We neglect negative value of time. The particle reaches origin of reference only once at t = 2 s. In order to determine time instant when particle is at initial position, we put x=10,

10 = 10 + 15 t + 1 2 X - 10 t 2 10 = 10 + 15 t + 1 2 X - 10 t 2 - t 2 + 3 t = 0 - t 2 + 3 t = 0 t = 0, 3 s t = 0, 3 s

The particle is at the initial position twice, including start point. Now, at the point of reversal of direction, speed of the particle is zero. Putting v = 0 and using v=u+at,

0 = 15 - 10 t 0 = 15 - 10 t t = 1.5 s t = 1.5 s

We neglect negative value and deduce that particle never ceases to move and as such there is no reversal of motion.

## Example

### Example 2

Problem : A particle’s velocity in “m/s” is given by a function in time “t” as :

v = 40 - 10 t v = 40 - 10 t

If the particle is at x = 0 at t = 0, find the time (s) when the particle is 60 m away from the initial position.

Solution : The velocity is given as a function in time “t”. Thus, we can know its acceleration by differentiating with respect to time :

a = đ v đ t = - 10 m / s 2 a = đ v đ t = - 10 m / s 2

Alternatively, we can see that expression of velocity has the form v = u + at. Evidently, the motion has a constant acceleration of -10 m / s 2 m / s 2 . Also, note that origin of reference and initial position are same. Applying equation of motion for position as :

x = u t + 1 2 a t 2 = 40 t + 1 2 x - 10 x t 2 = 40 t - 5 t 2 x = u t + 1 2 a t 2 = 40 t + 1 2 x - 10 x t 2 = 40 t - 5 t 2

According to question, we have to find the time when particle is 60 m away from the initial position. Since it is one dimensional motion, the particle can be 60 m away either in the positive direction of the reference or opposite to it. Considering that it is 60 m away in the positive reference direction, we have :

60 = 40 t - 5 t 2 60 = 40 t - 5 t 2

Re-arranging,

t 2 - 8 t + 12 = 0 t = 2 s or 6 s t 2 - 8 t + 12 = 0 t = 2 s or 6 s

We observe here that the particle is ultimately moving in the direction opposite to reference direction. As such it will again be 60 away from the initial position in the negative reference direction. For considering that the particle is 60 m away in the negative reference direction,

- 60 = 40 t - 5 t 2 - 60 = 40 t - 5 t 2

t 2 - 8 t - 12 = 0 t = - ( - 8 ) ± { ( - 8 ) 2 - 4 x 1 x ( - 12 ) } 2 = - 1.29 s , 9.29 s t 2 - 8 t - 12 = 0 t = - ( - 8 ) ± { ( - 8 ) 2 - 4 x 1 x ( - 12 ) } 2 = - 1.29 s , 9.29 s

Neglecting negative value of time,

t = 9.29 s t = 9.29 s

Note : We can also solve the quadratic equation for zero displacement to find the time for the particle to return to initial position. This time is found to be 8 seconds. We note here that particle takes 2 seconds to reach the linear distance of 60 m in the positive direction for the first time, whereas it takes only 9.29 – 8 = 1.29 second to reach 60 m from the initial position in the negative direction.

The particle is decelerated while moving in positive direction as velocity is positive, but acceleration is negative. On the other hand, both velocity and acceleration are negative while going away from the initial position in the negative direction and as such particle is accelerated. Therefore, the particle takes lesser time to travel same linear distance in the negative direction than in the positive direction from initial position..

Hence, the particle is 60 m away from the initial position at t = 2 s, 6 s and 9.29 s.


t (s)     0	 2	 4	 6	8        9.29
x (m)     0	60	80	60	0       -60
v (m/s)  40	20	 0     -20    -40       -52.9


### Example 3

Problem : A particle moves along x-axis with a velocity 9 m/s and acceleration -2 m / s 2 m / s 2 . Find the distance covered in 5 th 5 th second.

Solution : The displacement in 5 th 5 th second is :

x n = u + a 2 ( 2 n - 1 ) x n = 9 + - 2 2 ( 2 x 5 - 1 ) x n = 0 x n = u + a 2 ( 2 n - 1 ) x n = 9 + - 2 2 ( 2 x 5 - 1 ) x n = 0

The displacement in the 5 th 5 th second is zero. A zero displacement, however, does not mean that distance covered is zero. We can see here that particle is decelerated at the rate of -2 m / s 2 m / s 2 and as such there is reversal of direction when v = 0.

v = u + a t 2 = 9 - 2 t t = 4.5 s v = u + a t 2 = 9 - 2 t t = 4.5 s

This means that particle reverses its motion at t = 4.5 s i.e in the period when we are required to find distance. The particle, here, travels in the positive direction from t = 4 s to 4.5 s and then travels in the negative direction from t = 4.5 s to 5 s. In order to find the distance in 5 th 5 th second, we need to find displacement in each of these time intervals and then sum their magnitude to find the required distance.

The displacement for the period t = 0 s to 4 s is :

x = u t + 1 2 a t 2 = 9 x 4 + 1 2 x 2 x 4 2 = 36 - 16 = 20 m x = u t + 1 2 a t 2 = 9 x 4 + 1 2 x 2 x 4 2 = 36 - 16 = 20 m

The displacement for the period t = 0 s to 4.5 s is :

x = u t + 1 2 a t 2 = 9 x 4.5 + 1 2 x 2 x 4.5 2 = 40.5 - 20.25 = 0.25 m x = u t + 1 2 a t 2 = 9 x 4.5 + 1 2 x 2 x 4.5 2 = 40.5 - 20.25 = 0.25 m

The displacement for the period t = 4 s to 4.5 s is :

Δ x 1 = 20.25 - 20 = 0.25 m Δ x 1 = 20.25 - 20 = 0.25 m

Since particle is moving with constant acceleration in one dimension, it travels same distance on its return for the same period. It means that it travels 0.25 m in the period from t = 4.5 s to t = 5 s.

Δ x 2 = 0.25 m Δ x 2 = 0.25 m

Thus, total distance covered between t = 4 s to t = 5s i.e. in the 5 th 5 th second is :

Δ x = Δ x 2 + Δ x 2 = 0.25 + 0.25 = 0.5 m Δ x = Δ x 2 + Δ x 2 = 0.25 + 0.25 = 0.5 m

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