(a) By taking the slope of the curve in Figure 2.60, verify that the velocity of the jet car is\({\bf{115}}{\rm{ }}{\bf{m}}/{\bf{s}}\)at\(t = {\rm{ }}{\bf{20}}{\rm{ }}{\bf{s}}\). (b) By taking the slope of the curve at any point in Figure 2.61, verify that the jet car’s acceleration is\({\bf{5}}.{\bf{0}}{\rm{ }}{\bf{m}}/{{\bf{s}}^{\bf{2}}}\).

(a) Motion graphs can be used to analyse movement.

Graphical solutions for determining motion equations are identical to mathematical methods.

Velocity v is theslope of a graph of displacementx 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.

Here the acceleration can be calculated by obtaining the slop

\(\begin{array}{l}m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_2}}}\\Here\,the\,coordinates\,are\,(7,0)\,\,\,and\,\,\,(20,1500)\\m = \frac{{1500 - 0}}{{20 - 7}}\\m = 115.3\,m/s\end{array}\)

Hence the velocity is \(115.3 m/s\)

(b) Here if we carefully look into the figure 2.61, in your text book lets take any two points on the line

Coordinate (0, 17) and (30,160)

Slop of the line will be

\(\begin{array}{l}m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_2}}}\\Here\,the\,coordinates\,are\,(0,17)\,\,\,and\,\,\,(30,160)\\m = \frac{{160 - 17}}{{30 - 0}}\\m = 4.766\,m/{s^2}\end{array}\)

Hence the acceleration is approximately to the value \(5 m/{s^2}.\)

The velocity is \(115.3 m/s\). The acceleration is approximately to the value \(5 m/{s^2}.\)