The slope of a Velocity–time graph of an object moving in rectilinear motion with uniform velocity is straight line and parallel to x-axis when velocity is taken along y-axis and time is taken along x-axis.

Let two points A and B on the slope of graph.

Draw two lines parallel to y-axis AC from point A, and BD from point B.

Let point D at the x-axis (time axis) is t2 and point C is t1.

Let AB meet at ‘v’ at y-axis, i.e. object is moving with a velocity, v.

Thus, distance or displacement by the object is equal to the area of the rectangle (shaded) ABCD.

Thus, Area of `ABCD = BD xx DC`

`=> s = v (t_2 – t_1)`

Since given object is moving with constant velocity along a straight line, thus displacement will be equal to distance covered.

Therefore, Distance or Displacement = velocity X time interval.

When velocity – time graph is plotted for an object moving with uniform acceleration, the slope of the graph is a straight line.

The pattern of slope of the graph shows that object is moving with uniform acceleration.

Calculation of Displacement and Distance covered by the moving object using velocity time graph:

Let take two points, A and B at the slope of the graph.

Draw a line from B to BD and another from point A to AE parallel to y-axis.

Let AD meets at `t_2` and `AE` at `t_1` on the time axis.

Thus, Distance covered by the object in the given time interval `(t_2 – t_1)` is given by the area of ABCDE.

Therefore, Distance (s) = Area of `Delta ABCD +` Area of `ACDE`

`=>S=1/2 xx BC xx AC + (AExxED)`

Displacement of the object during the given time interval `(t_2-t_1)` = Area of ACDE

Thus, Displcement `= AE xx ED`

The slope of the velocity time graph of an object moving with uniform decreasing velocity with uniform acceleration is a downwards straight line. The straight downward slope shows the decreasing velocity with uniform acceleration, i.e. retardation.

Zig–zag line of slope of graph shows that the object is moving with non-uniform velocity.

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