Question

Given the velocity-time graph for a moving object, what is the displacement of the object during the 6.0 s of motion described in the graph?

Answer 1

The displacement of a velocity-time graph is given by the area under the curve.

To calculate the area under the curve, let us divide the area into two parts, as shown below.

We can calculate the area of the two shaded regions ( triangle and the rectangle) easily and add them up together.

The area of a triangle is given by,

`A_1=(1)/(2)* base* height`

From the diagram, base=6.0 s

And the height =50.0-20.0=30.0 m/s

On substituting the known values,

`\begin{gathered} A_1=(1)/(2)*6.0*30.0 \\ =90\text{ m} \end{gathered}`

The area of the shaded rectangle in the diagram is given by,

`A_2=\text{length }* width`

From the diagram, length = 6.0 s

And the width=20.0 m/s

On substituting the known values in the above equation,

`\begin{gathered} A_2=6.0*20.0 \\ =120\text{ m} \end{gathered}`

Thus the total displacement is given by,

`d=A_1+A_2`

On substituting the known values,

`\begin{gathered} d=90+120 \\ =210\text{ m} \end{gathered}`

Thus the displacement of the object during the 6.0 seconds of motion is 210 m