We next recall a general principle that will later be applied to distance-velocity-acceleration problems, among other things. If $F(u)$ is an anti-derivative of $f(u)$, then $\ds \int_a^bf(u)\,du=F(b)-F(a)$. Suppose that we want to let the upper limit of integration vary, i.e., we replace $b$ by some variable $x$. We think of $a$ as a fixed starting value $x_0$. In this new notation the last equation (after adding $F(a)$ to both sides) becomes: $$ F(x)=F(x_0)+\int_{x_0}^xf(u)\,du. $$ (Here $u$ is the variable of integration, called a "dummy variable,'' since it is not the variable in the function $F(x)$. In general, it is not a good idea to use the same letter as a variable of integration and as a limit of integration. That is, $\ds \int_{x_0}^xf(x)dx$ is bad notation, and can lead to errors and confusion.)
An important application of this principle occurs when we are interested in the position of an object at time $t$ (say, on the $x$-axis) and we know its position at time $\ds t_0$. Let $s(t)$ denote the position of the object at time $t$ (its distance from a reference point, such as the origin on the $x$-axis). Then the net change in position between $\ds t_0$ and $t$ is $\ds s(t)-s(t_0)$. Since $s(t)$ is an anti-derivative of the velocity function $v(t)$, we can write $$ s(t)=s(t_0)+\int_{t_0}^tv(u)du. $$ Similarly, since the velocity is an anti-derivative of the acceleration function $a(t)$, we have $$ v(t)=v(t_0)+\int_{t_0}^ta(u)du. $$
Example 9.2.1 Suppose an object is acted upon by a constant force $F$. Find $v(t)$ and $s(t)$. By Newton's law $F=ma$, so the acceleration is $F/m$, where $m$ is the mass of the object. Then we first have $$ v(t)=v(t_0)+\int_{t_0}^t{F\over m}\,du=v_0+ \left.{F\over m}u\right|_{t_0}^t=v_0+{F\over m}(t-t_0), $$ using the usual convention $\ds v_0=v(t_0)$. Then $$\eqalign{ s(t)&=s(t_0)+\int_{t_0}^t\left(v_0+{F\over m}(u-t_0)\right)du=s_0+ \left.(v_0u+{F\over2m}(u-t_0)^2)\right|_{t_0}^t\cr &=s_0+v_0(t-t_0)+{F\over2m}(t-t_0)^2.\cr }$$ For instance, when $F/m=-g$ is the constant of gravitational acceleration, then this is the falling body formula (if we neglect air resistance) familiar from elementary physics: $$s_0+v_0(t-t_0)-{g\over2}(t-t_0)^2,$$ or in the common case that $\ds t_0=0$, $$s_0+v_0t-{g\over2}t^2.$$ $\square$
Recall that the integral of the velocity function gives the net distance traveled, that is, the displacement. If you want to know the total distance traveled, you must find out where the velocity function crosses the $t$-axis, integrate separately over the time intervals when $v(t)$ is positive and when $v(t)$ is negative, and add up the absolute values of the different integrals. For example, if an object is thrown straight upward at 19.6 m/sec, its velocity function is $v(t)=-9.8t+19.6$, using $g=9.8$ m/sec$^2$ for the force of gravity. This is a straight line which is positive for $t< 2$ and negative for $t>2$. The net distance traveled in the first 4 seconds is thus $$\int_0^4(-9.8t+19.6)dt=0,$$ while the total distance traveled in the first 4 seconds is $$ \int_0^2(-9.8t+19.6)dt+\left|\int_2^4(-9.8t+19.6)dt\right|=19.6+|-19.6|=39.2 $$ meters, $19.6$ meters up and $19.6$ meters down.
Example 9.2.2 The acceleration of an object is given by $a(t)=\cos(\pi t)$, and its velocity at time $t=0$ is $1/(2\pi)$. Find both the net and the total distance traveled in the first 1.5 seconds.
We compute $$ v(t)=v(0)+\int_0^t\cos(\pi u)du={1\over 2\pi}+\left.{1\over\pi} \sin(\pi u)\right|_0^t={1\over\pi}\bigl({1\over2}+\sin(\pi t)\bigr). $$ The net distance traveled is then $$\eqalign{ s(3/2)-s(0)&=\int_0^{3/2}{1\over\pi}\left({1\over2}+\sin(\pi t)\right)\,dt\cr &=\left.{1\over\pi}\left({t\over2}-{1\over\pi}\cos(\pi t)\right) \right|_0^{3/2}={3\over4\pi}+{1\over\pi^2}\approx 0.340 \hbox{ meters.}\cr }$$ To find the total distance traveled, we need to know when $(0.5+\sin(\pi t))$ is positive and when it is negative. This function is 0 when $\sin(\pi t)$ is $-0.5$, i.e., when $\pi t=7\pi/6$, $11\pi/6$, etc. The value $\pi t=7\pi/6$, i.e., $t=7/6$, is the only value in the range $0\le t\le 1.5$. Since $v(t)>0$ for $t< 7/6$ and $v(t)< 0$ for $t>7/6$, the total distance traveled is $$\eqalign{ \int_0^{7/6}&{1\over \pi}\left({1\over2}+\sin(\pi t)\right)\,dt+ \Bigl|\int_{7/6}^{3/2} {1\over \pi}\left({1\over2}+\sin(\pi t)\right)\,dt\Bigr|\cr &={1\over \pi}\left( {7\over 12}+{1\over \pi}\cos(7\pi/6)+{1\over \pi}\right)+ {1\over \pi}\Bigl|{3\over 4}-{7\over 12} +{1\over \pi}\cos(7\pi/6)\Bigr|\cr &={1\over \pi}\left( {7\over 12}+{1\over \pi}{\sqrt3\over2}+{1\over \pi}\right)+ {1\over \pi}\Bigl|{3\over 4}-{7\over 12} +{1\over \pi}{\sqrt3\over2}.\Bigr| \approx 0.409 \hbox{ meters.}\cr }$$ $\square$
Exercises 9.2
For each velocity function find both the net distance and the total distance traveled during the indicated time interval (graph $v(t)$ to determine when it's positive and when it's negative):
Ex 9.2.1 $v=\cos(\pi t)$, $0\le t\le 2.5$ (answer)
Ex 9.2.2 $v=-9.8t+49$, $0\le t\le 10$ (answer)
Ex 9.2.3 $v=3(t-3)(t-1)$, $0\le t\le 5$ (answer)
Ex 9.2.4 $v=\sin(\pi t/3)-t$, $0\le t\le 1$ (answer)
Ex 9.2.5 An object is shot upwards from ground level with an initial velocity of 2 meters per second; it is subject only to the force of gravity (no air resistance). Find its maximum altitude and the time at which it hits the ground. (answer)
Ex 9.2.6 An object is shot upwards from ground level with an initial velocity of 3 meters per second; it is subject only to the force of gravity (no air resistance). Find its maximum altitude and the time at which it hits the ground. (answer)
Ex 9.2.7 An object is shot upwards from ground level with an initial velocity of 100 meters per second; it is subject only to the force of gravity (no air resistance). Find its maximum altitude and the time at which it hits the ground. (answer)
Ex 9.2.8 An object moves along a straight line with acceleration given by $a(t) = -\cos(t)$, and $s(0)=1$ and $v(0)=0$. Find the maximum distance the object travels from zero, and find its maximum speed. Describe the motion of the object. (answer)
Ex 9.2.9 An object moves along a straight line with acceleration given by $a(t) = \sin(\pi t)$. Assume that when $t=0$, $s(t)=v(t)=0$. Find $s(t)$, $v(t)$, and the maximum speed of the object. Describe the motion of the object. (answer)
Ex 9.2.10 An object moves along a straight line with acceleration given by $a(t) = 1+\sin(\pi t)$. Assume that when $t=0$, $s(t)=v(t)=0$. Find $s(t)$ and $v(t)$. (answer)
Ex 9.2.11 An object moves along a straight line with acceleration given by $a(t) = 1-\sin(\pi t)$. Assume that when $t=0$, $s(t)=v(t)=0$. Find $s(t)$ and $v(t)$. (answer)