Chapter 3, Problem 10STP

To determine

To Plot: The data in a graph.

To Calculate: Acceleration and displacement after $12.0\text{ s}$ .

Answer to Problem 10STP

The acceleration is $4.8\text{m}\cdot {\text{s}}^{\text{-2}}$ and the displacement $=444\text{m}$

Explanation of Solution

Introduction:

To plot a graph, the independent variable is plotted on the x- axis and the dependent variable on the y- axis.

Time is the independent variable. Hence, it is plotted on the x -axis while velocity is the dependent variable which is plotted on the y -axis. Acceleration can be calculated using the slope while the displacement can be calculated using the area of the graph.

  

The slope of the graph gives the acceleration.

The slope is the vertical change upon horizontal change.

  $slope=\frac{\Delta y}{\Delta x}$

  $slope=\frac{66-8}{12-0}$

Simplifying

  $slope=4.8\text{m}\cdot {\text{s}}^{-2}$

Hence, the acceleration is $4.8\text{m}\cdot {\text{s}}^{\text{-2}}$ .

To calculate the displacement, the area of the graph has to be calculated.

The area under the graph is the sum of the area of the rectangle and the area of the triangle.

Area of the rectangle $\text{=length×breadth}$

Area of triangle $=0.5\times \text{base×height}$

Total area $=\left(8\times 12\right)+0.5\left(12\times \left(66-8\right)\right)$

Total area $=444\text{m}$

Hence, the displacement $=444\text{m}$

Conclusion:

The acceleration is $4.8\text{m}\cdot {\text{s}}^{\text{-2}}$ and the displacement $=444\text{m}$ .