Chapter 2 - Differentiation - 2.2 Exercises - Page 116: 104

The graph of the corresponding position function is shown below.

Firstly divide the time into three intervals $[0,4]$, $[4,6]$ and $[6,10]$. In the interval $[0,4]$, velocity is constant and equal to $30$. Since velocity is constant, the position function will be a linear function with slope $=30$. At the time $t=0$, the position will also be zero. Thus $(0,0)$ will lie on the graph of the position function. Now use slope-intercept form to find the equation of the position function. We get, position function $d-0=30(t-0)$ or $d=30t$. Now calculate $d$ at time $t=4$. We get, $d=30\times4=120$ Thus, $(4,120)$ is a point on the graph of the position function of the interval $[4,6]$. Since, the velocity is zero in the interval $[4,6]$, the slope of the graph of position function will also be zero. Now again use slope-intercept form to find the equation of the position function. We get, $d-120=0(t-4)$ or $d=120$ Since the position is constant in the interval $[4,6]$. We get, $d=120$ at $t=6$. Thus, $(6,120)$ is a point on the graph of the position function of the interval $[6,10]$. Since, the velocity is constant and equals to $60$ in the interval $[6,10]$, the slope of the graph of position function will also be $60$. Again, use slope-intercept form to find the equation of the position function. We get, $d-120=60(t-6)$ or $d=60t-360+120=60t-240$ Now plot the three equations of the position function as follows.