91Ó°ÊÓ

The given data can be expressed below as:

This law states that an object continues to move in constant or uniform velocity unless an external force is acted upon on that object.

The equation of motion gives the distance moved and the displacement of the object along with the graph.

a) analysing from the graph, the area under the velocity curve is the distance travelled by the gazelle. The area of the trapezoid which has been formed in 0 s and 1 s is 80 cm. Moreover, the triangle from the 10 s - 12 s, has the area of 12 m.

Thus, the distance that the gazelle travels is 92 m which is the area under the curve.

b) As the gazelle has not run backwards, so it never had a negative velocity.

Thus, the displacement is the same for the gazelle as the distance travelled by the gazelle which is 92 cm.

c) The v - t graph’s slope is beneficial for identifying the acceleration of the gazelle.

The acceleration for the first is expressed as:

$\begin{array}{r}\text{acceleration}=\frac{\text{rise}}{\text{sun}}\\ =\frac{6\mathrm{m}/\mathrm{s}}{10\mathrm{s}}\\ =+0.6\mathrm{m}/{\mathrm{s}}^2\end{array}$

The acceleration for the final is expressed as:

$\begin{array}{r}\text{acceleration}=\frac{\text{rise}}{\text{sun}}\\ =\frac{-12\mathrm{m}/\mathrm{s}}{2\mathrm{s}}\\ =-6\mathrm{m}/{\mathrm{s}}^2\end{array}$

Analysing from the above data, the graph has been drawn below-