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Mathematics LibreTexts

9.2: Distance, Velocity, and Acceleration

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 $\ds t_0=0$, $$s_0+v_0t-{g\over2}t^2.$$



Recall that the integral of the velocity function gives the {\it net\/} distance traveled. If you want to know the {\it 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 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 {\it 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 {\it 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 $$\eqalign{ \int_0^{7/6}&{1\over \pi}\left({1\over2}+\sin(\pi t)\right)\,dt+ \Bigl|\int_{7/6}^{3/2} {1\over \pi}\left({1\over2}+\sin(\pi t)\right)\,dt\Bigr|\cr &={1\over \pi}\left( {7\over 12}+{1\over \pi}\cos(7\pi/6)+{1\over \pi}\right)+ {1\over \pi}\Bigl|{3\over 4}-{7\over 12} +{1\over \pi}\cos(7\pi/6)\Bigr|\cr &={1\over \pi}\left( {7\over 12}+{1\over \pi}{\sqrt3\over2}+{1\over \pi}\right)+ {1\over \pi}\Bigl|{3\over 4}-{7\over 12} +{1\over \pi}{\sqrt3\over2}.\Bigr| \approx 0.409 \hbox{ meters.}\cr }$$