§4.1 Determining distance traveled from velocity

Introduction

Area under the graph of the velocity function

Two approaches: area and antidifferentiation

When velocity is negative

A walker strolls at a constant $v(t) = 3$ miles per hour for two hours. The distance traveled equals the area under the velocity line on $[0,2]$, which is

A tossed ball has $v(t) = 32 - 32t$ feet per second and antiderivative $s(t) = 32t - 16t^2$ with $s(0) = 0$. The change in position $s(1) - s(\frac{1}{2})$ equals

For the same ball, $s(2) - s(0) = 0$. The correct interpretation is

A four-rectangle estimate with width $0.5$ gave $3.75$ miles for the walker's trip. To improve the estimate, the natural next step and its result are

With $v(t) = 0.5t^3 - 1.5t^2 + 1.5t + 1.5$, the position function satisfying $s' = v$ is

For the ball with $v(t) = 32 - 32t$, the antiderivative $s$ satisfying $s(0) = 0$ is

A velocity graph is positive on $0 \lt t \lt 1$, negative on $1 \lt t \lt 4$, and positive again on $4 \lt t \lt 8$. The position function

An object's velocity bounds areas of $1$ on $[0,1]$, $4$ on $[1,4]$, and $8$ on $[4,8]$, with the middle stretch below the axis. The total change in position and total distance traveled are

Plotting exact positions at integer times from velocity data, the position graph

With velocity data $v(0) = 1.5$, $v(0.5) = 1.9375$, $v(1.0) = 2$, $v(1.5) = 2.0625$ and rectangle width $0.5$ hours, compute the left endpoint estimate of distance on $[0,2]$, rounded to two decimals.

With $v(t) = 0.5t^3 - 1.5t^2 + 1.5t + 1.5$ and antiderivative $s(t) = \frac{1}{8}t^4 - \frac{1}{2}t^3 + \frac{3}{4}t^2 + \frac{3}{2}t$, compute $s(2) - s(0)$.

Between $t = \frac{1}{2}$ and $t = 1$ the ball's velocity line drops from $16$ to $0$ feet per second. Compute the exact area of the enclosed region.

An object's velocity bounds two triangular regions of area $1$ each on $[0,1]$ and $[1,2]$, with the second below the axis, and $s(0) = 1$. The total distance, net change, and position at $t = 2$ are

Using $s(t) = 32t - 16t^2$, compute $s(1) - s(0)$, the vertical distance traveled on $[0,1]$.

An object's position function is increasing on exactly those intervals where its velocity is positive.