Velocity and Acceleration (of an Object)

Direct Knowlege Contributor David A. Smith
  • By David A. Smith, Founder & CEO, Direct Knowledge
  • David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

We discuss the motion of an object in three dimensions. We study the path of an object, and its positions at various times, its velocity, speed and direction of travel of the object along the path. The acceleration at any time, and its effect on speed/velocity and the path.

Motion of an Object

If the graph of a vector function ${R}(t)$ is a smooth curve $C$ then the nonzero derivative ${R}'(t)$ is tangent to $C$ at the point $P$ that corresponds to $t$; and in this case we can make the following definition.

Definition. If an object moves in such a way that its position at time $t$ is given by the vector function ${R}(t)$ whose graph is a smooth curve $C,$ then

$(1) \quad {V(t)}=\frac{d {R}}{d t}$ is the object’s velocity vector at time $t$,

$(2) \quad |{V(t)}|$ is the speed of the object at time $t$,

$(3) \quad \frac{{V(t)}}{|{V(t)}|}$ is the direction of the object’s motion at time $t$, and

$(4) \quad \frac{d {V(t)}}{d t}$ is the object’s acceleration at time $t.$

Example. Suppose the position vector for a particle in space at time $t$ is given by $$ {R}(t)=e^t{i}+e^{-t}{j}+e^{2t}{k}. $$ Find the particle’s velocity vector, acceleration vector, speed, and direction of motion vector at time $t=\ln 2.$

Solution. At time $t=\ln 2$, the particle’s velocity vector is \begin{equation} {R}'(t)|{t=\ln 2}=e^t{i}-e^{-t}{j}+2e^{2t}{k}|{t=\ln 2}=2{i}-\frac{1}{2}{j}+8{k}, \end{equation} the acceleration vector is \begin{equation} {R}”(t)|{t=\ln 2}=e^t{i}+e^{-t} {j} + 4e^{2t} {k} |{t=\ln 2}=2{i}+\frac{1}{2}{j}+16{k}, \end{equation} the speed is \begin{equation} {R}'(t)|_{t=\ln 2} = \sqrt{2^2+\left(\frac{-1}{2}\right)^2+8^2}=\sqrt{68.25}, \end{equation} and the direction of motion vector is \begin{equation} \frac{2}{\sqrt{68.25}}{i}-\frac{1}{ 2\sqrt{68.25 }}{j}+\frac{8}{\sqrt{68.25}}{k}. \end{equation} as desired.

Example. Find the position vector ${R}(t)$ and velocity vector ${V}(t),$ given the acceleration vector function $$ {A}(t)=t^2 {i}-2\sqrt{t } {j}+e^{3t} {k}, $$ initial position vector ${R}(0)=2 {i}+{j}-{k},$ and initial velocity vector ${V}(0)={i}-{j}-2{k}.$

Solution. Given ${A}(t)=t^2 {i}-2\sqrt{t } {j}+e^{3t} {k}$ the velocity vector function is \begin{equation} {V}(t)=\left(\int t^2 dt\right){i}-\left(\int 2\sqrt{t }dt\right) {j}+\left(\int e^{3t} dt\right) {k} \end{equation} \begin{equation} {V}(t)=\left(\frac{t^3}{3}+C_1\right) {i}-\left(\frac{4 t^{3/2}}{3}+C_2\right){j}+\left(\frac{e^{3 t}}{3}+C_3\right){k} \end{equation} where $C_1, C_2,$ and $C_3$ are constants to be determined. By using \begin{equation}{V}(0)={i}-{j}-2{k} =\left(0+C_1\right){i}+\left(-0+C_2\right){j}+\left(\frac{1}{3}+C_3\right){k} \end{equation} we find $C_1=1,$ $C_2=1,$ $C_3=-7/3.$ Therefore, \begin{equation} {V}(t)=\left(\frac{t^3}{3}+1\right){i}+\left(-\frac{4 t^{3/2}}{3}-1\right){j}+\left(\frac{e^{3 t}-7}{3}\right) {k}. \end{equation} So the position vector function is \begin{equation} {R}(t)=\left(\int \left(\frac{t^3}{3}+1\right) \, dt\right) {i}+\left(\int \left(-\frac{4 t^{3/2}}{3}-1\right) \, dt\right){j}+\left(\int \frac{e^{3 t}-7}{3} \, dt\right) {k} \end{equation} which is
\begin{equation} {R}(t)=\left(\frac{t^4}{12}+t+K_1\right) {i}+\left(-\frac{8 t^{5/2}}{15}-t+K_2\right){j}+\left(\frac{1}{3} \left(\frac{e^{3 t}}{3 }-7 t\right)+k_3\right) {k} \end{equation} where $K_1, K_2,$ and $K_3$ are constants to be determined. By using \begin{equation} {R}(0)=2 {i}+{j}-{k} =\left(0+K_1\right){i}+\left(-0+K_2\right){j}+\left(\frac{1}{9} + K_3 \right){k} \end{equation} we find $K_1=2,$ $K_2=1,$ $K_3=-10/9.$ Therefore, $${R}(t)=\left(\frac{t^4}{12}+t+2\right) {i}+\left(-\frac{8 t^{5/2}}{15}-t+1\right){j}+\left(\frac{1}{3} \left(\frac{e^{3 t}}{3 }-7 t\right)-\frac{10}{9}\right) {k} $$ is the required vector function.

The Path of a Projectile

Theorem. Neglecting air resistance, the path of a projectile launched from an initial height $h$ with initial speed $v_0$ and an angle of elevation $\theta$ is described by the vector function \begin{equation} \label{prolaueq} {R}(t) =(v_0\cos \theta)t{i}+\left[h+(v_0\sin \theta)t-\frac{1}{2}g t^2)\right]{j} \end{equation} where $g$ is the gravitational constant.

Proof. Suppose a projectile of mass $m$ is launched from an initial position ${R}_0$ with an initial velocity ${V}_0.$ First we will find its position vector as a function of time. We begin with acceleration ${A}(t)=-g {j}$ and we integrate twice, namely, \begin{align} & {V}(t)=\int {A}(t)\, dt=\int -g {j} \, dt =-g t {j}+{C}_1 \\ & {R}(t)= \int {V}(t) \, dt =\int (g t {j}+{C}_1) \, dt =-\frac{1}{2} g t^2 {i}+{C}_1 t+{C}_2 \end{align} where ${C}_1$ and ${C}_2$ can be determined from the initial conditions. Using ${V}(0)={V}_0$, ${R}(0)={R}_0$, produces ${C}_1={V}_0$ and ${C}_2={R}_0.$ Therefore the position vector function is \begin{equation} \label{proeq} {R}(t)=-\frac{1}{2} g t^2 {j} +t {V}_0 +{R}_0. \end{equation} Now from the given height $h$ we realize that $${R}_0=h {j} $$ and because speed is the magnitude of the velocity, that is $V_0=||{V}_0||$, we have $$ {V}_0=x{i}+y {j} =\left(||{V}_0|| \cos \theta \right) {i}+\left(||{V}_0|| \sin \theta \right) {j} =V_0\cos \theta{i}+V_0\sin \theta{j}. $$ Now we can derive \eqref{prolaueq} as follows \begin{align} R(t) & =-\frac{1}{2} g t^2 {j} +t {V}_0+{R}_0 \\ & = -\frac{1}{2} g t^2 {j} +tV_0 \cos \theta {i} + t (V_0\cos \theta{i}+V_0\sin \theta{j}) +h {j} \\ & =(v_0\cos \theta)t{i}+\left[h+(v_0\sin \theta)t-\frac{1}{2}g t^2)\right]{j} \end{align} as desired.

Exercises on Velocity and Acceleration

Exercise. Find the unit tangent vector ${T}(t)$ for ${r}(t) = 8t{i}+8t{j}+4t{k}$ at $t=2.$

Exercise. Find the unit tangent vector ${T}(t)$ and the unit normal vector ${N}(t).$

$(1) \quad {R}(t)=t^2{i}+\sqrt{t}{j},$ with $t>0$

$(2) \quad {R}(t)=(t \cos t){i}+(t \sin t){j}$

$(3) \quad {R}(t)=t{i}+(\ln \cos t){j}, \, -\pi/20$

$(4) \quad {R}(t)=(\cos^3 t){i}+(\sin^3 t){j}+{k}, \, 0<t<\pi/2$

Exercise. Given ${R}(t)=(\ln t){i}+t^2{j}$ find $a$ so that
\begin{equation} {N}(t)=\frac{-2}{\sqrt{1+a t^a}}{i}+\frac{1}{\sqrt{1+a t^a}}{j}.\end{equation}

Exercise. Given ${R}(t)=(\cos t){i}+(\sin t){j}+t {k}$ find $a$ so that ${N}(t)=(- \cos a t) {i}+(-\sin a t){j}.$

Exercise. Determine values for $a$ and $b$ so that the following vector functions are smooth over $[a,b].$

$(1) \quad {F}(t)=a t^3{i}+b t^2{j}+ a b t {k}$

$(2) \quad {G}(t)=(a \sin b t){i}+(a \cos b t){j}+ a b {k}$

$(3) \quad {H}(t)=\left(e^{a t}-b t\right){i}+t^{a/b}{j}+ (b \cos a t) {k}$

Exercise. The velocity of a particle moving in space is ${V}(t)=e^t{i}+t^2{j}.$ Find the vector ${R}(0)$ so that the particle’s position as a function of $t$ is \begin{equation} {R}(t)=e^t{i}+\left(\frac{1}{3}t^3-1\right){j}. \end{equation}

Exercise. The acceleration of a moving particle is ${A}(t)=24t^2{i}+4 {j}.$ Find the vectors ${R}(0)$ and ${V}(0)=0$ so that the particle’s position as a function of $t$ is \begin{equation} {R}(t)=\left(2t^4+1\right){i}+\left(2t^2+2\right){j}. \end{equation}

Exercise. Show that ${N}(t)=-g'(t){i}+f'(t){j}$ and $-{N}(t)=g'(t){i}-f'(t){j}$ are both normal to the curve ${R}(t)+f(t){i}+g(t){j}$ at the point $(f(t),g(t)).$

Exercise. A baseball is hit above ground at 100 feet per second at an angle of $\pi/4$ with respect to the ground. Find the maximum height reached by the baseball. Will it clear a 10-foot high fence located 300 feet from home plate?

Exercise. Find the vector function for the path of a projectile launched at a height of 10 feet above the ground with an initial velocity of 88 feet per second and at an angle of $30^\circ$ above the horizontal. Sketch the path of the projectile.

Exercise. Find the angle at which an object must be thrown to obtain (a) the maximum range, and (b) he maximum m height.

Exercise. A projectile is fired from ground level at an angle of $10^\circ$ with the horizontal. Find the minimum initial velocity necessary of the projectile is to have a range of 100 feet.

Exercise. Use linear approximation by finding a set of parametric equations for the tangent line to the graph at $t=t_0$ and use the equations for the line to approximate ${R}(t_0+.01).$

$(1) \quad {R}(t)=t {i}-t^2 {j} +\frac{t^3}{4}$, $t_0=1$

$(2) \quad {R}(t)=t {i}+\sqrt{25-t^2} {j} +\sqrt{25-t}$, $t_0=3$