MA005: Calculus I

In Fig. 14, $\mathrm{f}(t)$ is the velocity of a car in "miles per hour", and t is the time in "hours". Then the shaded "area" will be $\text { (base) } \cdot \text { (height })=(3 \text { hours }) \cdot(20 \text { miles } / \text { hour })=60 \text { miles }$, the distance traveled by the car in the 3 hours from 1 o'clock until 4 o'clock.

Distance as an "Area"

If $\mathrm{f}(t)$ is the (positive) forward velocity of an object at time $t$, then the "area" between the graph of $f$ and the t–axis and the vertical lines at times $t=\mathrm{a}$ and $t=\mathrm{b}$ will be the distance that the object has moved forward between times $a$ and $b$.

This "area as distance" can make some difficult distance problems much easier.

Example 3: A car starts from rest (velocity = 0) and steadily speeds up so that 20 seconds later it's speed is 88 feet per second (60 miles per hour). How far did the car travel during those 20 seconds?

Solution: We can answer the question using the techniques of chapter 3 (try it). But if "steadily" means that the velocity increases linearly, then it is easier to use Fig. 15 and the idea of "area as distance". The "area" of the triangular region represents the distance traveled, so

$\begin{aligned} \text { distance } &=\frac{1}{2}(\text { base }) \cdot(\text { height })=\frac{1}{2}(20 \text { seconds }) \cdot(88 \text { feet } / \text { second }) \\ &=880 \text { feet. } \end{aligned}$.

Practice 7: A train traveling at 45 miles per hour (66 feet/second) takes 60 seconds to come to a complete stop. If the train slowed down at a steady rate (the velocity decreased linearly), how many feet did the train travel while coming to a stop?

Practice 8: You and a friend start off at noon and walk in the same direction along the same path at the rates shown in Fig. 16. (a) Who is walking faster at 2 pm? Who is ahead at 2 pm? (b) Who is walking faster at 3 pm? Who is ahead at 3 pm? (c) When will you and your friend be together? (Answer in words.)