Answer to Question #246261 in Calculus for Jess

Question

11. Suppose a car travels at a constant rate of 70 kph for 2 hours and travels 55 kph
thereafter. Show that distance travelled is a function of time and find the rule of the
function.
12. A man jogs for 20 minutes at a rate of 6 kph, then walks for 30 minutes at a rate of 3 kph,
then sits and rests for 15 minutes and finally walks for 45 minutes. Find the rule of the
function that expresses his distance travelled as a function of time

Answer 1

11. $Distance=Speed \times Time$

Let $d(t)$ be the distance function. Where $t$ is the time.

$d(t)=70\times 2+55\times t\\ d(t)=55t+140$

It is clear that distance is a function of time $t$ .

12. For the first 20 minutes, $d(t)=6t$ when $0\leq t \le\frac{1}{3}$

Next 30 minutes, $d(t)=2+3(t-\frac{1}{3})$ when $\frac{1}{3}<t\leq \frac{5}{6}$

Next 15 minutes, $d(t)=3.5$ when $\frac{5}{6}< t\leq \frac{13}{12}$

Last 45 minutes, $d(t)=3.5+3(t-\frac{13}{12})$ when $\frac{13}{12}<t\leq \frac{11}{6}$

Therefore,

$d(t)=\begin{cases} 6t ~~~~~~~~~~~~~~~~~~~~~~~~~~\text{ when } 0\leq t \leq\frac{1}{3}\\ 2+3(t-\frac{1}{3}) ~~~~~~~~\text{ when } \frac{1}{3}<t\leq \frac{5}{6}\\\frac{5}{6} 3.5~~~~~~~~~~~~~~~~~~~~~~~~\text{ when } \frac{5}{6}< t\leq \frac{13}{12}\\ 3.5+3(t-\frac{13}{12}) ~~~~~\text{ when } \frac{13}{12}<t\leq \frac{11}{6} \end{cases}$