Q61P - University Physics with Modern Physics

Question

A gazelle is running in a straight line (the x-axis). The graph in Fig. P2.61 shows this animal’s velocity as a function of time. During the first 12.0 s, find (a) the total distance moved and (b) the displacement of the gazelle. (c) Sketch an ${\mathit{a}}_{\mathbf{x}}\mathbf{-}\mathit{t}$ graph showing this gazelle’s acceleration as a function of time for the first 12.0 s.

Step-by-Step Solution

Answer

a) the distance that the gazelle travels are 92 m which is the area under the curve, b) the displacement is the same for the gazelle as the distance travelled by the gazelle which is 92 m and c) the $a_x-t$ graph has been illustrated below.

Step 1: Identification of the given data

The given data can be expressed below as:

• The time taken by the gazelle to run in a straight line is 12.0 s.

Step 2: Significance of the Newton’s first law of motion for the gazelle

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.

Step 3: Determination of the distance moved, displacement and 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-