Contents
- 1 Nature of the Distance – Time Graphs for Uniform and Non-Uniform motion of a Body
- 1.1 1. Distance-Time Graphs
- 1.2 2. Speed-Time Graphs (Or Velocity-Time Graphs)
- 1.3 (i) Speed-Time Graph when the Speed Remains Constant
- 1.4 (ii) Speed-Time Graph when Speed Changes at a Uniform Rate (Uniform Acceleration)
- 1.5 Speed-Time Graph when the Initial Speed of the Body is Not Zero
- 1.6 (iii) Speed-Time Graph when Speed Changes at a Non-Uniform Rate (Non-Uniform Acceleration)
- 1.7 Displacement-time Graph
- 1.8 Velocity-time Graph
The laws of Physics Topics are used to explain everything from the smallest subatomic particles to the largest galaxies.
Nature of the Distance – Time Graphs for Uniform and Non-Uniform motion of a Body
We will now discuss the various types of graphs which can be used to calculate ‘speed’ (or velocity), ‘acceleration’ and ‘distance travelled’ by a body. A very important point to remember here is that in the drawing of graphs, the terms ‘speed’ and ‘velocity’ are used in the same sense. In most of the graphs which we are going to discuss now, we will be using the term ‘speed’.
At some of the places, however, the term ‘velocity’ will also be used. This is because even in the examination papers, they use both the terms ‘speed’ as well as ‘velocity’ in the questions based on graphs. We will first study the ‘distance-time graphs’ and then ‘speed-time graphs’ (or velocity-time graphs). Please note that in drawing graphs based on motion, ‘time’ is always taken along the x-axis whereas ‘distance’ or ‘speed’ (or velocity) is taken along the y-axis.
1. Distance-Time Graphs
When a body moves with uniform speed, it will travel equal distances in equal intervals of time. In other words, the distance travelled is directly proportional to time. Thus, for uniform speed, a graph of distance travelled against time will be a straight line as shown by line OA in Figure.
Please note that we can also write the term “uniform velocity” in place of “uniform speed” in the graph. We can now say that the distance-time graph of a body moving at uniform speed is always a straight line. In other words, a straight line graph between distance and time tells us that the body is moving with a uniform speed.
The slope of a distance-time graph indicates speed of the body. So, the distance-time graph of a body can be used to calculate the speed of the body. To find the speed from the distance-time graph of a body, we take any point A on the straight line graph (Figure), and drop a perpendicular AB on the time axis (x – axis). It is clear that AB represents the distance travelled by the body in the time interval represented by OB (Figure). We know that:
Speed = \(\frac{\text { Distance travelled }}{\text { Time taken }}\)
Now, in Figure, the distance travelled is OC which is in fact equal to AB, and the time taken is OB. So, putting distance = AB and time = OB in the above relation, we get:
Speed = \(\frac{A B}{O B}\)
But \(\frac{A B}{O B}\) is known as the slope (or gradient) of the graph line OA, therefore, in a distance-time graph for uniform speed, the speed of the body is given by the slope of the graph. In other words, the slope of a distance-time graph indicates speed.
Thus, we can use a distance-time graph of a body to find the value of speed of the body. All that we have to do is to find out the slope of distance-time graph as shown above and that gives the speed of the body.
We have seen above that if the speed of a body is uniform then its distance-time graph is a straight line (as shown in Figure ). If, however, the speed of a body is non-uniform, then the graph between distance travelled and time is a curved line (called a parabola) as shown in Figure.
We get the following conclusions from the above discussion :
- If the distance-time graph of a body is a straight line, then its speed is uniform.
- If the distance-time graph of a body is a curved line, then its speed is non-uniform.
We know that when a body moves with a non-uniform speed, then its motion is said to be accelerated. So, the curved line OA in Figure also represents the distance-time graph of a body moving with accelerated motion.
If a displacement-time graph is drawn, then it will specifically represent ‘velocity’. For example, if the displacement-time graph of a moving body is a straight line, then it represents uniform velocity of the body.
We will now discuss the speed-time graphs of a moving body. In fact, the speed-time graphs are also known as velocity-time graphs. So, whether we write a speed-time graph or a velocity-time graph, it will mean the same thing.
2. Speed-Time Graphs (Or Velocity-Time Graphs)
We can have three types of speed-time graphs for a moving body. These three cases an
- When the speed of the body remains constant (and there is no acceleration)
- When the speed of the body changes at a uniform rate (there is uniform acceleration
- When the speed of the body changes in a non-uniform way (there is non-uniform acceleration)
We will now discuss these three types of speed-time graphs in detail, one by one.
(i) Speed-Time Graph when the Speed Remains Constant
By saying that the speed of a body is constant (or uniform), we mean that the speed does not change with time and hence there is no acceleration. So, a speed-time graph for a body moving with constant speed (or uniform speed) is a straight line parallel to the time axis (as shown by the line AB in Figure).
In other words, if the speed-time graph of a body is a straight line parallel to the time axis, then the speed of the body is constant (or uniform). Since the speed of the body is constant or uniform, there is no acceleration, and hence there is no question of finding the acceleration from such a speed-time graph. We can, however, find the distance travelled by the body in a given time from such a speed-time graph. This is described below.
We know that, Speed = \(\frac{\text { Distance travelled }}{\text { Time taken }}\)
So, Distance travelled = Speed × Time taken ………….. (1)
Now, to find out the distance travelled by the body at point C (Figure 35), we draw a perpendicular CB at point C which meets the straight-line graph at point B.
Now, Speed at C = CB
But CB = OA
Thus, Speed at C = OA …….. (2)
And, Time at C = OC ……… (3)
Now, putting these values of speed and time in relation (1), we get:
Distance travelled = OA × OC (see Figure)
or Distance travelled = Area of rectangle OABC
Thus, in a speed-time graph, the area enclosed by the speed-time curve and the time axis gives us the distance travelled by the body.
It should be noted that we can also use the term “velocity” in place of “speed” everywhere in the above discussion. Thus, we can say that the velocity-time graph of an object moving with constant velocity (or uniform velocity) is a straight line parallel to the time axis.
We can also say that in a velocity-time graph, the area enclosed by the velocity-time curve and the time axis gives the distance travelled by the object. We will now discuss the speed-time graph of a body when its speed changes at a uniform rate, that is, when the acceleration of the body is uniform.
(ii) Speed-Time Graph when Speed Changes at a Uniform Rate (Uniform Acceleration)
When a body moves with uniform acceleration, its speed changes by equal amounts in equal intervals of time. In other words, the speed becomes directly proportional to time. Thus, the speed-time graph for a uniformly changing speed (or uniform acceleration) will be a straight line (as shown by line OP in Figure).
We can find out the value of acceleration from the speed-time graph of a moving body. Now, to calculate the acceleration at a time corresponding to point Q (Figure), we draw a perpendicular QP from point Q which touches the straight line graph at point P. We know that:
Acceleration = \(\frac{\text { Change in speed (or velocity) }}{\text { Time taken }}\)
In Figure, the change in speed is represented by PQ whereas time taken is equal to OQ. So,
Acceleration = \(\frac{P Q}{O Q}\)
But \(\frac{P Q}{O Q}\) is the slope (or gradient) of the speed-time graph OP, therefore, we conclude that in a speed-time graph, the acceleration is given by the slope of the graph. In other words, the slope of a speed-time graph of a moving body gives its acceleration.
The distance travelled by a moving body in a given time can also be calculated from its speed-time graph. As explained earlier, the distance travelled by the body in the time corresponding to point Q (Figure) will be equal to the area of the triangle OPQ, which is. equal to half the area of the rectangle ORPQ. Thus, Distance travelled = Area of triangle OPQ
= \(\frac{1}{2}\) Area of rectangle ORPQ
= \(\frac{1}{2}\) × OR × OQ (see Figure)
Please note that in a speed-time graph of a body, a straight line sloping upwards (as in Figure) shows uniform acceleration. On the other hand, in a speed-time graph of a body, a straight line sloping downwards (as in Figure) indicates uniform retardation.
It should be noted that we can also use the word “velocity” in place of “speed” everywhere in the above discussion. Thus, we can also say that the velocity-time graph of a body moving with uniform acceleration is a straight line. And that the slope of velocity-time graph of a body indicates its acceleration.
In the speed-time graph shown in Figure, we have assumed that the initial speed of the body is zero. It is, however, possible that the body has some initial speed and then it starts accelerating at a uniform rate. So, we will now discuss the speed-time graph of a body whose initial speed is not zero.
Speed-Time Graph when the Initial Speed of the Body is Not Zero
Figure shows the speed-time graph of a body having an initial speed equal to OB and then accelerating from B to C. In order to calculate the value of acceleration from such a graph, we will have to subtract the initial speed (OB) from the final speed (AC), and then divide it by time (OA).
In such cases also, the distance travelled by the body in a given time is equal to the area between the speed-time graph and the time axis. For example, in this case the distance travelled by the body in time OA (Figure), will be equal to the area of the figure OBCA under the speed-time graph BC.
Now, the figure OBCA has two parallel sides OB and AC and such a figure is known as a trapezium. Thus, the distance travelled by the body in this case is equal to the area of trapezium OBCA. Now,
Area of trapezium = \(\frac{\text { Sum of two parallel sides } \times \text { Height }}{2}\)
Here, sum of parallel sides is OB + AC and height is OA (see Figure).
So, Distance travelled = \(\frac{(O B+A C) \times O A}{2}\)
We will now discuss the speed-time graph of a body whose speed does not change at a uniform rate, that is, when the acceleration of the body is non-uniform.
(iii) Speed-Time Graph when Speed Changes at a Non-Uniform Rate (Non-Uniform Acceleration)
When the speed of a body changes in an irregular manner, then the speed-time graph of the body is a curved line (as shown by the line OA in- Figure). Even now, the distance travelled by the body is given by the area between the speed-time curve and the time axis.
Once again, please note that we can write the word “velocity” in place of “speed” in the above graph. So, we can also say that the velocity-time graph for non-uniform acceleration is a curved line called parabola. We will now solve some problems based on graphs.
Example Problem 1.
Study the speed-time graph of a body given here and answer the following questions :
(a) What type of motion is represented by OA ?
(b) What type of motion is represented by AB ?
(c) What type of motion is represented by BC ?
(d) Find out the acceleration of the body.
(e) Calculate the retardation of the body.
(f) Find out the distance travelled by the body from A to B.
Solution:
(a) OA is a straight line graph between speed and time, and it is sloping upwards from O to A. Therefore, the graph line OA represents uniform acceleration.
(b) AB is a straight line graph between speed and time, which is parallel to the time axis (v-axis). So, AB represents uniform speed (or constant speed). There is no acceleration from A to B.
(c) BC is a straight line graph between speed and time which is sloping downwards from B to C. Therefore, BC represents uniform retardation (or negative acceleration).
(d) Let us find out the acceleration now. We have just seen that the graph line OA represents acceleration. So, the slope of speed-time graph OA will give us the acceleration of the body. Thus,
Acceleration = Slope of line OA
= \(\frac{A D}{O D}\)
Now, in the given graph (Figure), we find that AD = 6 m/s and OD = 4 seconds. So, putting these values in the above relation, we get:
Acceleration = \(\frac{6 \mathrm{~m} / \mathrm{s}}{4 \mathrm{~s}}\)
= 1.5 m/s2 ……………. (1)
(e) Let us calculate the retardation now. We have discussed above that the graph line BC represents retardation. So, the slope of speed-time graph BC will be equal to the retardation of the body. So,
Retardation = Slope of line BC
= \(\frac{B E}{E C}\)
Now, in the graph given to us (Figure), we find that BE = 6 m/s and EC = 16 – 10 = 6 seconds. So, putting these values in the above relation, we get:
Retardation = \(\frac{6 \mathrm{~m} / \mathrm{s}}{6 \mathrm{~s}}\)
= 1 m/s2 ………….. (2)
(f) We will now find out the distance travelled by the body in moving from A to B (Figure). We have studied that in a speed-time graph, the distance travelled by the body is equal to the area enclosed between the speed-time graph and the time-axis. Thus,
Distance travelled from A to B = Area under the line AB and the time axis
= Area of rectangle DABE
= DA × DE
Now, from the given graph (Figure), we find that DA = 6 m/s and DE – 10 – 4 = 6 s. Therefore,
Distance travelled from A to B = 6 × 6
= 36 m ……………….. (3)
Here is an Exercise for You : Find (i) Distance travelled from O to A, (ii) Distance travelled from B to C, and (iii) Total distance travelled by the body. The answers will be 12 m, 18 m, and 66 m respectively. For this purpose you will require the formula for the area of a triangle. Please note that:
Area of a triangle = \(\frac{1}{2}\) × base × height
A yet another point to be noted is that in the graph given in the above sample problem (Figure 40) they could also have written the word “velocity” in place of “speed”.
Example Problem 2.
A car is moving on a straight road with uniform acceleration. The following table gives the speed of the car at various instants of time :
Draw the speed-time graph by choosing a convenient scale. Determine from it:
(i) the acceleration of the car.
(ii) the distance travelled by the car in 50 seconds.
Solution:
We take a graph paper and plot the above given time values on the .Y-axis. The corresponding speed values are plotted on the y-axis. The speed-time graph obtained from the given readings is shown in Figure. Please note that in this case, when the time is 0, then the speed is not 0. The body has an initial speed of 5 m/s which is represented by point A in Figure. We will now answer the questions asked in this sample problem.
(i) Calculation of Acceleration. We know that :
Acceleration = Slope of speed-time graph
= Slope of line AF (see Figure )
= \(\frac{F G}{A G}\)
Now, if we look at the graph shown in Figure, we will find that the value of speed at point F is 30 m/s and that at point G is 5 m/s.
Therefore, FG = 30 – 5
= 25 m/s
Again, at point G, the value of time is 50 seconds whereas that at point A is 0 second.
Thus, AG = 50 – 0
= 50 s
Now, putting these values of FG and AG in the above relation, we get:
Acceleration = \(\frac{25 \mathrm{~m} / \mathrm{s}}{50 \mathrm{~s}}\)
= 0.5 m/s2
(ii) Calculation of Distance Travelled. The distance travelled by the car in 50 seconds is equal to the area under the speed-time curve AF. That is, the distance travelled is equal to the area of the figure OAFH (see Figure). But the figure OAFH is a trapezium. So,
Distance travelled = Area of trapezium OAFH
= \(\frac{(\text { Sum of two parallel sides }) \times \text { Height }}{2}\)
In Figure, the two parallel sides are OA and HF whereas the height is OH. Therefore,
Distance travelled = \(\frac{(O A+H F) \times O H}{2}\)
= \(\frac{(5+30) \times 50}{2}\)
= \(\frac{35 \times 50}{2}\)
= 875 m
Displacement-time Graph
A graph obtained by plotting t (time) along the x-axis and s, the corresponding distances travelled by a particle along the y-axis is called distance-time graph. When the corresponding distances (s) are plotted as displacements along the y -axis, the graph is called displacement-time graph. These graphs represent the changes in position and, hence, the displacement of a particle with time.
Figures are representative graphs for a particle
- at rest [Fig.],
- in motion with uniform velocity[Fig.],
- in motion with a uniform acceleration [Fig.] and
- in motion with non-uniform acceleration [Fig.].
In Fig., the point P denotes the displacement OR of the particle in time OQ. In this figure, the gradient of the displacement-time graph determines the velocity of the particle. The straight line has a uniform gradient—so the velocity is uniform.
According to Fig., the particle is displaced by s2 – s1 = CB in the time interval t2 – t1 = AC.
Therefore, the average velocity of the particle in that interval of time, v = \(\frac{s_2-s_1}{t_2-t_1}\) = \(\frac{C B}{A C}\) = gradient of the chord AB.
To find the instantaneous velocity of the particle at time t1, the time interval (t2 – t1) needs to be infinitesimally small. Hence, the point B is almost superimposed on point A. In this condition, the gradient of the chord AB becomes equal to the gradient of the tangent drawn at A. Thus, the gradient of the tangent drawn at any point on the displacement-time graph denotes the instantaneous velocity of the particle at the corresponding moment of time.
Velocity-time Graph
A velocity-time graph is drawn by plotting time t along the x-axis and velocity v along the y-axis. Figures show velocity-time graphs for a particle
- moving with a uniform velocity [Fig.],
- starting from rest and moving with a uniform acceleration [Fig.],
- starting with an initial velocity u and accelerating uniformly [Fig.] and
- in motion with non-uniform acceleration [Fig.].
In figures, the point B denotes the magnitude of the velocity OC of the particle in time OA. In these figures, the gradient of the velocity-time graph gives the acceleration of the particle. In Fig., the gradient of CB is zero—so there is no acceleration. But in Fig., the gradient of OB is positive and uniform—so the acceleration is uniform.
In figures, the area under the velocity-time graph and the time-axis gives the displacement of the particle.
In Fig., the average acceleration of the particle in the time interval (t2 – t1) is equal to the gradient of the chord AB. The graph denotes the motion of a particle moving with non-uniform acceleration.
With the help of calculus it can be shown that, for a particle moving with a non-uniform acceleration, its displacement for any interval of time is equal to the area enclosed by the arc denoting the motion, the time interval and the time-axis.