# Acceleration – Time Graph of Various Types of Motions of a Particle

## Acceleration – Time Graph of Various Types of Motions of a Particle

Let the acceleration be given as the function of time, i.e. a = f (t). Then the elementary change in velocity during a time dt is $$d\overrightarrow{v}=\overrightarrow{a}dt$$. We can observe that in a – t graph, $$\overrightarrow{a}dt$$ is equal to the area of the shaded elementary strip.

Summing up all the elementary areas dA, we have the total area $$A=\int{dA=\int{\overrightarrow{a}\,}}dt$$. Since, $$\int\limits_{{{t}_{1}}}^{{{t}_{2}}}{\overrightarrow{a}\,dt}=\int\limits_{{{v}_{1}}}^{{{v}_{2}}}{\overrightarrow{dv}}=\Delta \overrightarrow{v}$$; We have $$\Delta \overrightarrow{V}=\int{\overrightarrow{a}}\,dt$$ = A = Area under a – t graph. That means, the change in velocity $$\Delta \overrightarrow{v}$$ over any time interval is represented by the algebraic sum of all positive and negative area which we call area under a – t graph over that time interval, $$A=\sum{{{A}_{+}}}+\sum{{{A}_{-}}}$$. If the area is negative, $$\Delta \overrightarrow{v}$$ is negative, $$\Delta \overrightarrow{v}$$ is directly in negative direction of coordinate axes. When area is zero, $$\Delta \overrightarrow{v}=0$$, which means no change in velocity.

$$\Delta \overrightarrow{V}=\int{\overrightarrow{a}}\,dt$$ = Area under v – t graph; $${{A}_{+}}\to \Delta \overrightarrow{{{v}_{1}}}$$ is + ve; $${{A}_{-}}\to \Delta \overrightarrow{{{v}_{2}}}$$ is – ve.

When $$\Delta \overrightarrow{V}$$ is positive, you should not immediately accept it as $$\Delta |\overrightarrow{V}|>0$$, because $$\Delta \overrightarrow{V}$$ is a vector quantity. Hence, positive area means a positive $$\Delta \overrightarrow{v}$$ but not positive $$\Delta \overrightarrow{v}\,(\ne \Delta V)$$.