\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \)
\( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \)
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17calculus > vector functions > projectile motion

Vector Functions - Projectile Motion

This page covers the basics of working with the position, velocity and acceleration vector functions. The acceleration vector that you learn about on this page can be expressed in terms of the unit tangent vector and the principal unit normal vector, which you can find on the acceleration vector components page.

Calculating The Velocity and Acceleration

Projectile motion using vector functions works just as you would expect. The following table lists the equations. If \(\vec{r}(t)\) is a vector function describing the position of a projectile, the velocity is \(\vec{v}(t)=\vec{r}'(t)\) and the acceleration is \(\vec{a}(t)=\vec{v}'(t)=\vec{r}''(t)\).

Position

\(\vec{r}(t) =x(t)\vhat{i}+y(t)\vhat{j}+z(t)\vhat{k}\)

Velocity

\(\vec{v}(t)=x'(t)\vhat{i}+y'(t)\vhat{j}+z'(t)\vhat{k}\) \(=\vec{r}'(t) \)

Speed

\(\| \vec{v}(t) \|\)

Acceleration

\(\vec{a}(t)=x''(t)\vhat{i}+y''(t)\vhat{j}+z''(t)\vhat{k}\) \(=\vec{v}'(t)\) \(=\vec{r}''(t)\)

There are really no surprises here. Notice that the speed is just the magnitude of the velocity and so it's value is always a positive scalar. Some instructors use the terms speed and velocity interchangeably but they actually refer to different things.

Calculating The Position Vector

Sometimes we are given the acceleration vector or the velocity vector and asked to calculate the position vector. In those cases we use integration. As you would expect, to get the velocity vector from the acceleration vector, we use these equations.

Acceleration

\(\vec{a}(t)=a_x(t)\vhat{i}+a_y(t)\vhat{j}+a_z(t)\vhat{k}\)

Velocity

\(\vec{v}(t)=\int{a_x(t)~dt}\vhat{i} + \int{a_y(t)~dt}\vhat{j} + \int{a_z(t)~dt}\vhat{k} + \vec{C} =\) \(\int{\vec{a}(t)~dt} + \vec{C}\)

Note - - In the equation for the velocity vector, the vector \(\vec{C}\) is the constant vector that we get when we do integration. The velocity vector could also be written \(\vec{v}(t)=\left[ \int{a_x(t)~dt} + C_x \right] \vhat{i} + \left[ \int{a_y(t)~dt} + C_y \right] \vhat{j} + \left[ \int{a_z(t)~dt} + C_z \right] \vhat{k} \) where \(\vec{C} = C_x\vhat{i} + C_y\vhat{j} + C_z\vhat{k}\).

Once we have the velocity vector (or if we are given the velocity vector), we can calculate the position vector. For the equations below, we assume the velocity vector is in the form \(\vec{v}(t) = v_x\vhat{i} + v_y\vhat{j} + v_z\vhat{k}\).

Velocity

\(\vec{v}(t) = v_x\vhat{i} + v_y\vhat{j} + v_z\vhat{k}\)

Position

\(\vec{r}(t) = \int{v_x~dt}\vhat{i} + \int{v_y~dt}\vhat{j} + \int{v_z~dt}\vhat{k} + \vec{K}\)

In each of the above equations we end up with general constants, in our case \(\vec{C}\) and \(\vec{K}\). You will probably run across problems that give you information that you can use to find the actual values of these constants. Most of the time the information is given in the form of initial conditions, i.e. values of velocity and/or position at time \(t=0\). However, values at any other time will also allow you to find the constants. To do this, you substitute the value for time into the final equation and evaluate. Some practice problems demonstrate how to do this.

Acceleration Vector Components

The acceleration vector \(\vec{a}(t)= x''(t)\vhat{i}+y''(t)\vhat{j}+z''(t)\vhat{k}\) can be expressed in terms of two other unit vectors, the unit tangent vector and the principal unit normal vector. After working some practice problems, you need to learn how to calculate the other two unit vectors before learning how to write the acceleration using them.

next: unit tangent vector →

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Practice Problems

Instructions - - Unless otherwise instructed, give your answers in exact form.

Level A - Basic

Practice A01

For the position function \(\vec{r}(t)=\langle t-\sin t,1-\cos t\rangle\), find the velocity and acceleration vectors at the point \((\pi,2)\).

answer

solution

Practice A02

Find the position vector function \(\vec{r}(t)\) for a particle with acceleration \(\vec{a}(t) = \langle 2t,2\sin(t),\cos(4t)\rangle\), initial velocity \(\vec{v}(0)=\langle 1,-3,2\rangle\) and initial position \(\vec{r}(0)=\langle 2,4,-1\rangle\).

answer

solution

Practice A03

Determine the velocity vector, speed and acceleration vector of an object when \(t=1\) given by the position vector \(\vec{r}(t)=t\vhat{i}+(-0.5t^2+4)\vhat{j}\).

answer

solution

Practice A04

Determine the velocity vector, speed and acceleration vector of an object when \(t=2\) for the position vector \(\vec{r}(t)=\cos(\pi t)\vhat{i}+\sin(\pi t)\vhat{j}+(t^2/2)\vhat{k}\)

answer

solution

Practice A05

Find the velocity and acceleration vectors and speed of the vector function \(\vec{r}(t)=\langle -5t,-3t^2,4t^4+3\rangle\) at \(t=1\).

answer

solution

Practice A06

A car travels with a velocity vector given by \(\vec{v}(t)=\langle t^2,e^t+1\rangle\), where t is measured in seconds and the vector components are measured in feet. If the initial position of the car is \(\vec{r}(0)=\langle 1,3 \rangle\), find the position of the car after one second.

answer

solution

Practice A07

A particle moves along a curve whose parametric equations are \(x(t)=e^{-t}\), \(y(t)=2\cos(3t)\), \(z(t)=2\sin(3t)\). (a) Determine the velocity and acceleration vectors.
(b) Find the magnitude of the velocity (speed) and acceleration at \(t=0\).

answer

solution

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