Q69P - Fundamentals Of Physics

Question

How far does the runner whose velocity time graph is shown in Fig 2-40 travel in$\mathbf{16}\mathbf{ }\mathbf{s}$?The figure’s vertical scaling is set by ${\mathbf{V}}_{\mathbf{s}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{m}\mathbf{/}\mathbf{s}$.

Step-by-Step Solution

Answer

The distance covered by the runner in$16s$ is $100m$.

Step 1: Given data

The plot of $v\left(m/s\right)$against $\mathrm{t}\left(\mathrm{s}\right)$with vertical scale is set by $v_s=8\left(m/s\right)$.

Step 2: Area under the curve

The calculation of the region bounded by the curve within the given graph is called the area under the curve.There are four different shapes of the area under the curve. The first shape is a triangle, the second shape is a rectangle. The third shape is a triangle at the upper part and a rectangle at the lower part, and the fourth shape is a rectangle. 

Step 3: Calculations for distance traveled by runner

Therefore, the distance covered within 16s will beequal to the total area under the curve. It is equal to the area of the shapes under the velocity curve.

 

It is calculated using the given values in the problem as below:

$$\begin{gathered} \mathrm{X}=\left(\frac{1}{2}\times 2\mathrm{s}\times 8\mathrm{m}/\mathrm{s}\right)+\left(8\mathrm{s}\times 8\mathrm{m}/\mathrm{s}\right)+\left(\left(2\mathrm{s}\times 4\mathrm{m}/\mathrm{s}\right)+\frac{1}{2}\times 2\mathrm{s}\times 4\mathrm{m}/\mathrm{s}\right)+\left(4\mathrm{s}\times 4\mathrm{m}/\mathrm{s}\right) \\ =100\mathrm{m} \end{gathered}$$

Hence, the distance traveled by a runner is $100\mathrm{m}$.