64 PE - College Physics (Urone)

Question

a) Take the slope of the curve in Figure 2.64 to find the jogger’s velocity at t = 2.5 s. (b) Repeat at 7.5 s. These values must be consistent with the graph in Figure 2.65.

Step-by-Step Solution

Answer

The velocity of the graph at 2.5 time is approx 3.57 m/s

Creating graph and its analysis

Motion graphs can be used to analyze movement.

For determining motion equations, the mathematical procedures and graphical solutions are the same.

The velocity v is represented by the slope of a graph of displacement x vs. time t.

Acceleration is the inclination of a graph of velocity v vs. time t.

Graphs can be used to calculate average velocity, instantaneous velocity, and acceleration.

Calculating velocity at 2.5 s

 Here if we carefully look into the figure, in your text book lets take any two points on the line 

Coordinate \(\left( {0,0} \right)\) and \(\left( {7,25} \right)\) 

Slop of the line will be

\[\begin{array}{l}m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_2}}}\\Here\,the\,coordinates\,are\,(0,0)\,\,\,and\,\,\,(7,25)\\m = \frac{{25 - 0}}{{7 - 0}}\\m = 3.57\,m/s\end{array}\]
 

The velocity of the graph at 2.5 time is \[3.57 m/s\] approx.

Calculating velocity at 7.5 s

Here if we carefully look into the figure, in your text book lets take any two points on the line 

Coordinate \(\left( {11,0} \right)\)  and \(\left( {4,25} \right)\) 

Slop of the line will be

\[\begin{array}{l}m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_2}}}\\Here\,the\,coordinates\,are\,(11,0)\,\,\,and\,\,\,(4,25)\\m = \frac{{25 - 0}}{{4 - 11}}\\m =  - 3.57\,\,m/s\end{array}\]
 

The velocity of the graph at 7.5 s time is \[3.57 m/s\] approx.

Hence the value which we derived also consist the figure 2.64.