# JEE Main & Advanced Physics One Dimensional Motion Velocity-time Graph

Velocity-time Graph

Category : JEE Main & Advanced

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

Calculation of Distance and displacement : The area covered between the velocity time graph and time axis gives the displacement and distance travelled by the body for a given time interval.

Total distance $=\,|{{A}_{1}}|\,+\,|{{A}_{2}}|\,+\,|{{A}_{3}}|\,$

= Addition of modulus of different area. i.e. $s\,=\,\int{|\upsilon |\,dt}$

Total displacement $={{A}_{1}}+{{A}_{2}}+{{A}_{3}}$

= Addition of different area considering their sign.

i.e. $r\,=\,\int{\upsilon \,dt}$

Area above time axis is taken as positive, while area below time axis is taken as negative

here ${{A}_{1}}$ and ${{A}_{2}}$ are area of triangle 1 and 2 respectively and ${{A}_{3}}$ is the area of trapezium

Calculation of Acceleration : Let AB is a velocity-time graph for any moving particle

As Acceleration = $\frac{\text{Change in velocity}}{\text{Time taken}}$

$=\frac{{{v}_{2}}-{{v}_{1}}}{{{t}_{2}}-{{t}_{1}}}$   ...(i)

From triangle ABC, $\tan \theta =\frac{BC}{AC}=\frac{AD}{AC}$

$=\frac{{{v}_{2}}-{{v}_{1}}}{{{t}_{2}}-{{t}_{1}}}$   ...(ii)

By comparing (i) and (ii)

Acceleration (a) = $\tan \theta$

It is clear that slope of tangent on velocity-time graph represents the acceleration of the particle.

Various velocity -time graphs and their interpretation

 $\theta ={{0}^{o}},\,\,a=0,\,\,\upsilon =$ constant i.e., line parallel to time axis represents that the particle is moving with constant velocity. $\theta ={{0}^{o}},\,\,a=\infty ,\,\,\upsilon =$ increasing i.e., line perpendicular to time axis represents that the particle is increasing its velocity, but time does not change. It means the particle possesses infinite acceleration. Practically it is not possible. $\theta =$ constant, so  a = constant and $\upsilon$ is increasing uniformly with time i.e., line with constant slope represents uniform acceleration of the particle. $\theta$ increasing so acceleration increasing i.e., line bending towards velocity axis represent the increasing acceleration in the body. $\theta$ decreasing so acceleration decreasing i.e. line bending towards time axis represents the decreasing acceleration in the body Positive constant acceleration because $\theta$ is constant and $<{{90}^{o}}$ but initial velocity of the particle is negative. Positive constant acceleration because $\theta$ is constant and $<{{90}^{o}}$ but initial velocity of particle is positive. Negative constant acceleration because $\theta$ is constant and $>{{90}^{o}}$ but initial velocity of the particle is positive. Negative constant acceleration because $\theta$ is constant and $>{{90}^{o}}$ but initial velocity of the particle is zero. Negative constant acceleration because $\theta$ is constant and $>{{90}^{o}}$ but initial velocity of the particle is negative.

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