NEET Physics Motion in a Straight Line / सरल रेखा में गति Position Time Graph

Position Time Graph            

During motion of the particle its parameters of kinematical analysis (u, v, a, r) changes with time. This can      be represented on the graph.

Position time graph is plotted by taking time t along x-axis and position of the particle on y-axis.

Let AB is a position-time graph for any moving particle

As Velocity \[=\frac{\text{Change in position}}{\text{Time taken}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{t}_{2}}-{{t}_{1}}}\]                       … (i)

From triangle ABC \[\tan \theta =\frac{BC}{AC}=\frac{AD}{AC}=\frac{{{y}_{2}}-{{y}_{1}}}{{{t}_{2}}-{{t}_{1}}}\]           …. (ii)

By comparing (i) and (ii)         \[\operatorname{Velocity} = tan\,\theta \]

\[v = tan\,\theta \]

It is clear that slope of position-time graph represents the velocity of the particle.

Various position – time graphs and their interpretation

	q = 0o so v = 0 i.e., line parallel to time axis represents that the particle is at rest.
	q = 90o so v = ¥ i.e., line perpendicular to time axis represents that particle is changing its position but time does not changes it means the particle possesses infinite velocity. Practically this is not possible.
	q = constant so v = constant, a = 0 i.e., line with constant slope represents uniform velocity of the particle.
	\[\theta \] is increasing so v is increasing, a is positive. i.e., line bending towards position axis represents increasing velocity of particle. It means the particle possesses acceleration.
	q is decreasing so v is decreasing, a is negative i.e., line bending towards time axis represents decreasing velocity of the particle. It means the particle possesses retardation.
	\[\theta \,\,constant but \,>\, 9{{0}^{o}}\] so v will be constant but negative i.e., line with negative slope represent that particle returns towards the point of reference. (negative displacement).
	Straight line segments of different slopes represent that velocity of the body changes after certain interval of time.
	This graph shows that at one instant the particle has two positions. Which is not possible.
	The graph shows that particle coming towards origin initially and after that it is moving away from origin.

Note: If the graph is plotted between distance and time then it is always an increasing curve and it             never comes back towards origin because distance never decrease with time. Hence such type of distance           time graph is valid up to point A only, after point A it is not valid as shown in the figure.

For two particles having displacement time graph with slopes \[{{\theta }_{1}}\,and\,\,{{\theta }_{2}}\] possesses velocities \[{{v}_{1}}\,and\text{ }{{v}_{2}}\]     respectively then \[\frac{{{\upsilon }_{1}}}{{{\upsilon }_{2}}}\,=\,\frac{\tan {{\theta }_{1}}}{\tan {{\theta }_{2}}}\]

Sample problems based on position-time graph

Problem 20. The position of a particle moving along the x-axis at certain times is given below:

Which of the following describes the motion correctly [AMU (Engg.) 2001]

(a) Uniform, accelerated               (b) Uniform, decelerated

(c) Non-uniform, accelerated         (d) There is not enough data for generalisation

Solution: (a) Instantaneous velocity \[v=\frac{\Delta x}{\Delta t}\], By using the data from the table

\[{{v}_{1}}=\frac{0-(-2)}{1}\,=\,2m/s\,\,,\,{{v}_{2}}=\,\frac{6-0}{1}\,=\,\,6m/s\] and \[{{v}_{3}}=\,\frac{16-6}{1}\,=\,\,10\,m/s\] i.e. the speed is increasing at a constant rate so            motion is uniformly accelerated.

Problem 21. Which of the following graph represents uniform motion [DCE 1999]

           (a)                    (b)                  (c)        (d)            

Solution: (a) When distance time graph is a straight line with constant slope than motion is uniform.

Problem 22. The displacement-time graph for two particles A and B are straight lines inclined at angles of             \[{{30}^{o}}and\text{ }{{60}^{o}}\]with the time axis. The ratio of velocities of \[{{\operatorname{v}}_{A}}: \,{{v}_{B}}\] is [CPMT 1990; MP PET 1999; MP PET 2001]

(a) 1 : 2             (b) \[1:\sqrt{3}\] (c) \[\sqrt{3}:1\]  (d) 1 : 3

Solution: (d) \[v=\tan \theta \] from displacement graph. So \[\frac{{{v}_{A}}}{{{v}_{B}}}=\frac{\tan {{30}^{o}}}{\tan {{60}^{o}}}=\frac{1/\sqrt{3}}{\sqrt{3}}=\frac{1}{\sqrt{3}\times \sqrt{3}}=\frac{1}{3}\]

Problem 23.     From the following displacement time graph find out the velocity of a moving body

(a) \[\frac{1}{\sqrt{3}}\,m/s\]                              (b) \[3 m/s\]

(c) \[\sqrt{3}\,m/s\]                                 (d) \[\frac{1}{3}\]

Solution: (c) In first instant you will apply \[\upsilon =\tan \theta \] and say, \[\upsilon =\tan {{30}^{o}}=\frac{1}{\sqrt{3}}\,m/s\].

But it is wrong because formula \[\upsilon =\tan \theta \] is valid when angle is measured with time axis.

Here angle is taken from displacement axis. So angle from time axis \[={{90}^{o}}-{{30}^{o}}={{60}^{o}}\].

Now \[\upsilon =\tan \,{{60}^{o}}=\sqrt{3}\]

Problem 24.     The diagram shows the displacement-time graph for a particle moving in a straight line. The average             velocity for the interval t = 0, t = 5 is

(a) 0                                          (b) \[6 m{{s}^{1}}\]

(c) \[ 2 m{{s}^{1}}\]                           (d) \[2\text{ }m{{s}^{1}}\]

Solution: (c) Average velocity \[=\frac{\text{Total}\,\,\text{displacement }}{\text{Total}\,\,\text{time}}=\frac{\left( 20 \right)+\left( -20 \right)+\left( -10 \right)}{5}~=\,2\text{ }m/s\]

Problem 25. Figure shows the displacement time graph of a body. What is the ratio of the speed in the first second             and that in the next two seconds

(a) 1 : 2                         (b) 1 : 3

(c) 3 : 1                                     (d) 2 : 1

Solution: (d) Speed in first second = 30 and Speed in next two seconds = 15. So that ratio 2 : 1