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The shape of the graph gives a clear idea o f the nature of variation of one quantity with respect to the other i.e., it provides us the relationship between the two quantities represented by it. For example, if the graph is a straight line and passes through the origin, it is of the form = or if the graph is a straight line having positive or negative intercepts on y-axis, it is represented by, = ± . From the slope of the graph and its intercepts on the x or y-axis, we can calculate the values of physical quantities or universal constants. For example, In photoelectric effect experiment, versus υ graph comes out to be a straight straight line, giving an intercept intercept on x-axis as well as y-axis. Slope of this straight line graph provides the value of Planck’s constant while the intercepts on y and x-axes give the values of work f unction and threshold frequency respectively. Inaccuracies in the experimental data can also be identified with the help of a graph. With the help of a graph, we can also find out the mean value from a large number of observations.
(i) Straight line graph
(a) Passing through origin : This graph is represented by the equation = . Graph of this nature is obtained when we plot two quantities which are proportional to each other, e.g. velocity-time graph of a body starting from rest and having a uniform acceleration (Fig.1). (b) Not passing through origin : This graph is represented by the equation = + . Such type of graph is obtained when we plot two quantities which are proportional to each other, but one is not zero when the other is, e.g. velocity-time graph of a body having initial velocity and uniform acceleration (Fig. 2). (c) Having negative slope : Such graph points downwards with the line making an obtuse angle with the x-axis, e.g. velocity-time graph of a body projected vertically upwards till it reaches the maximum height (Fig. 3.)
(ii) Parabola
a) Symmetric parabola : This graph is represented by the equation = 2 . Such type of graph is obtained when one quantity varies directly as the square of the other. The graph passes through the origin and is bisected by one of the axes equally into two symmetrical portions e.g. Kinetic energy and momentum of a body ( 2 = 2) (Fig. 4) b) Asymmetric parabola : This graph is represented by the equation = + 2 . This also passes through the origin and only one portion of graph is obtained at a time, e.g. distance-time graph of body starting with an initial velocity and traveling with uniform acceleration, equation : distance = + (1/2) 2 . (Fig.5). (iii) Rectangular Hyperbola
This graph is represented by the equation = 2 . If we plot graph between two quantities in inverse proportion, we get a rectangular hyperbola e.g. pressure and volume of an ideal gas at constant temperature, equation : = . (Fig. 6) (iv) Exponential
When two quantities are related by an exponential equation, we get this graph, equation : = − e.g. graph between the number of atoms remaining to undergo radioactive disintegration and time (Fig. 7). (v) Other graphs
(a) When a quantity decreases and then increases : The graph of angle of incidence and angle of deviation produced by a prism is an example of this case (Fig. 8). (b) When a quantity increases and remains constant : The graph of velocity and time, of a body starting with uniform acceleration and attaining uniform velocity (Fig. 9) is a very good example of t his case. (c) When a quantity decreases and remains constant : The graph of speed and time, of a body starting with an initial speed and uniform retardation, reaching a constant final speed, is a good example of this case (Fig. 10).
a) The slope of displacement – time graph gives the velocity. If the graph is a straight line parallel to time axis then the body is at rest. When t he time is noted it has some initial displacement.
b) The two graphs “b” and “c” represent a body moving with constant velocity having some initial displacement and zero initial displacement respectively.
c) The graph is a parabola. It is for a body starting from rest and moving with uniform acceleration (i.e., increasing slope)
d) The graph is a parabola. It shown that the body is moves with retardation ( is decreasing slope)
e) The graph is a parabola. It shows that the body is moving with uniform acceleration. (since slope increases).
f)
The body is moving is opposite direction to the initial direction [since with increasing time displacement decreases.]
g) The graph is a parabola. It shows that the body moves with retardation up to = 1 then it moves with acceleration up to = 2 and reaches the initial point. Here 1 = 2 = This shows a body projected vertically upwards.
The slope of velocity time graph acceleration and the area under the velocity time graph gives the displacement. a) This graph is a straight line parallel to time axis represent a body moving with uniform velocity.
b) This graph is a straight line passing through origin and represents a body moving with uniform acceleration (i.e., slope is constant and positive) starting from rest.
c) This graph is a straight line having some intercept on velocity axis and represents a body moving with uniform acceleration starts with an initial velocity.
d) This graph is a parabola represents a body moving with increasing acceleration (i.e. slope increases)
e) This graph represents a body moving with uniform retardation starting with an initial velocity. The body comes to rest when = 1
f)
The velocity – time graph of a ball freely falling from a height and undergoing elastic collisions with the floor is as shown in fig f (consider velocity is positive during upward motion and is negative during downward motion) OA, QC, SE : Descent. BQ, DS : Ascent OP, QR : time of descent, PQ, RS : Time of ascent. AB, CD during collision with the floor only direction of velocity changes without change in magnitude.
g) This graph represents the inelastic collision with the floor. After successive collisions, the ball comes to rest.
h) The graph represents a body projected vertically upwards from the ground Here OA = DC = Magnitude of velocity
∆ le OAB + A rea of ∆ OB = time of ascent BD = time of descent Here distance traveled by the body = Area of le BCD = 2 Area of ∆ OAB Displacement traveled by the body = Area of ∆ OAB – Area of ∆ BCD = 0
i.
The graph represents a body projected vertically upwards from the top of a tower [From fig i] OA = Velocity with which it is projected upwards [+ ] = Velocity with which it crosses the top of a tower downwards. [−] = Velocity of body when it strikes the ground. [ − 1 ] The total distance traveled by the body = ∆ + area of ∆ Displacement of the body = ∆ – area of ∆ = − Height of tower =
a.
This graph represents a body moving with uniform acceleration.
b.
The graph represents a body moving with non-uniform acceleration. The area of ( − ) graph with time axis is equal to change in velocity.
