Answer
a. $15$. $3.33$. $10$ mi/hr.
b. $10,0,4$ mi/hr
c. $3.5$ hr, $20$ mi/hr
(Other estimates are possible.)
Work Step by Step
a. Reading the function values from the graph, we get $D(0)=0, D(1)=15, D(2.5)=20, D(3.5)=30$. The average speed over the time interval $[0,1]$ is given by $S_1=\frac{D(1)-D(0)}{1-0}=15-0=15$ mi/hr. For interval $[1,2.5]$, we have $S_2=\frac{D(2.5)-D(1)}{2.5-1}=\frac{20-15}{1.5}\approx 3.33$ mi/hr. And for interval $[2.5,3.5]$, we have $S_3=\frac{D(3.5)-D(2.5)}{3.5-2.5}=\frac{30-20}{1}=10$ mi/hr.
b. The instantaneous speed at time $t$ is the slope of the curve at that time, and we can estimate from the graph that: $S(\frac{1}{2})=10,S(2)=0, S(3)=4$ (draw tangent lines at each point and measure the slope).
c. The maximum speed happened when the slope of the curve reaches a maximum (or the steepest point) which is at $t=3.5$ hr and gives $S(3.5)=20$ mi/hr
