# 9.2: Distance, Velocity, and Acceleration

- Page ID
- 487

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

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 \(\int_a^bf(u)\,du=F(b)-F(a)\). 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, \(\int_{x_0}^xf(x)dx\) 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 \( t_0\). 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 \(t_0\) and \(t\) is \(s(t)-s(t_0)\). 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 $ 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 \(t_0=0\), \(s_0+v_0t-{g\over2}t^2.\)

Recall that the integral of the velocity function gives the *net* distance traveled. 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 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.

**Solution**

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 $$\eqalign{ \int_0^{7/6}&{1\over \pi}\left({1\over2}+\sin(\pi t)\right)\,dt+ \Bigl|\int_{7/6}^{3/2} {1\over \pi}\left({1\over2}+\sin(\pi t)\right)\,dt\Bigr|\cr &={1\over \pi}\left( {7\over 12}+{1\over \pi}\cos(7\pi/6)+{1\over \pi}\right)+ {1\over \pi}\Bigl|{3\over 4}-{7\over 12} +{1\over \pi}\cos(7\pi/6)\Bigr|\cr &={1\over \pi}\left( {7\over 12}+{1\over \pi}{\sqrt3\over2}+{1\over \pi}\right)+ {1\over \pi}\Bigl|{3\over 4}-{7\over 12} +{1\over \pi}{\sqrt3\over2}.\Bigr| \approx 0.409 \hbox{ meters.}\cr }$$