Math 344: Calculus III

13.4 Velocity and Acceleration

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13.4.1 Velocity and Speed

In this section, we will discuss kinematics , the study of bodies in motion. Specifically, we want to talk about the position , velocity , and acceleration of a particle in 3D space. We begin by considering the position vector $\mathbf{r}(t)$ of a particle. In one-dimensional motion, we know from previous courses that the velocity $\mathbf{v}(t)$ of a particle is given by taking the derivative with respect to $t$, $\mathbf{v}(t) = \mathbf{r}'(t)$. This holds for 3D motion as well, with

$$ \def\ihat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ı}}}} \def\jhat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ȷ}}}} \def\khat{\mathbf{\hat{k}}} \mathbf{v}(t) = \lim_{\Delta t\rightarrow 0} \dfrac{\mathbf{r}(t+\Delta t) - \mathbf{r}(t)}{\Delta t} = \mathbf{r}'(t) $$
Therefore, the velocity vector is equivalent to the tangent vector for 3D motion and points in the same direction as the tangent line to the space curve.

Speed is the magnitude of the velocity $\left|\mathbf{v}(t)\right|$. From last section , we know that this is also related to arc length $s$, since velocity is the change of position with respect to time along the curve. Hence,

$$ \left|\mathbf{v}(t)\right| = \left|\mathbf{r}'(t)\right| = \dfrac{ds}{dt} $$

13.4.2 Acceleration

In the one-dimensional case, we know that acceleration is the derivative of the velocity with respect to time, so

$$ \mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{r}''(t) $$
This formula holds for space curves, hence the vector notation, but we have to make sure we understand what is meant here. When we say "one-dimensional" motion in this context, we mean with respect to the parameter $t$ (or $s$ if we parametrize with respect to arc length). That is velocity and acceleration are computed respect to the path that the particle is traveling on. If you are traveling on a train, the only possible directions is can move are forward and backward on the track. The track causes it to go north, south, east or west (or some combination of these) controlling your $x$ and $y$ motion. The track may also have elevation changes, controlling your $z$ motion. The train does not directly control any of these changes in direction, as it can only go forward or backward along the track. This is the "one-dimension" you travel in, along the path.

Exercise 13.4.1

Find the position, acceleration, and speed of a particle whose position function is

$$ \mathbf{r}(t) = t\,\ihat + t^2\,\jhat + 2\,\khat $$
Find the velocity and acceleration vectors at $t = 1$, then sketch the path of the particle and these two vectors.

Check Your Work

$$ \mathbf{v}(t) = \mathbf{r}'(t) = \ihat + 2t\,\jhat $$
$$ \mathbf{a}(t) = \mathbf{v}'(t) = 2\,\jhat $$
$$ \left|\mathbf{v}(t)\right| = \sqrt{(1)^2 + (2t)^2} = \sqrt{1 + 4t^2} $$
To make the sketch, we need to find the velocity and acceleration at $t =1$

$$ \mathbf{v}(1) = \ihat + 2\,\jhat \qquad\qquad \mathbf{a}(1) = 2\,\jhat $$

Example 13.4.1

Given the acceleration function

$$ \mathbf{a}(t) = 2t\,\ihat + \sin t\,\jhat + \cos 2t\,\khat $$
and the initial values $\mathbf{v}(0) = \ihat$ and $\mathbf{r}(0) = \jhat$, find the position vector function for a particle and sketch the particle's path.

Solution

Since $\mathbf{v}'(t) = \mathbf{a}(t)$, we can of course use the Fundamental Theorem of Calculus to integrate $\mathbf{a}(t)$ to find $\mathbf{v}(t)$.

$$ \begin{align*} \mathbf{v}(t) &= \int \mathbf{a}(t)\,dt \\ \\ &= \int \left(2t\,\ihat + \sin t\,\jhat + \cos 2t\,\khat\right)\,dt \\ \\ &= t^2\,\ihat - \cos t\,\jhat + \frac{1}{2}\sin 2t\,\khat + \mathbf{C} \end{align*} $$
Here $\mathbf{C}$ is the constant of integration. To find its value, we use the condition that $\mathbf{v}(0) = \ihat$, so

$$ \begin{align*} \mathbf{v}(0) = \ihat &= 0\,\ihat - \cos(0)\,\jhat + \frac{1}{2}\sin 2(0)\,\khat + \mathbf{C} \\ \\ &= -\jhat + \mathbf{C} \\ \\ \mathbf{C} &= \ihat + \jhat \end{align*} $$
Hence $\mathbf{v}(t) = (t^2+1)\,\ihat + (1 - \cos t)\,\jhat + \frac{1}{2}\sin 2t\,\khat$. We use this to find the position function

$$ \begin{align*} \mathbf{r}(t) &= \int \mathbf{v}(t)\,dt \\ \\ &= \int \left((t^2+1)\,\ihat + (1 - \cos t)\,\jhat + \frac{1}{2}\sin 2t\,\khat\right)\,dt \\ \\ &= \left(\frac{t^3}{3} + t\right)\,\ihat + \left(t - \sin t\right)\,\jhat - \frac{1}{4}\cos 2t\,\khat + \mathbf{K} \end{align*} $$
Once, more we use an initial condition $\mathbf{r}(0) = \jhat$ to find the value of the constant of integration.

$$ \begin{align*} \mathbf{r}(0) &= \jhat = \left(\frac{0^3}{3} + 0\right)\,\ihat + \left(0 - \sin 0\right)\,\jhat - \frac{1}{4}\cos 2(0)\,\khat + \mathbf{D} \\ \\ &= -\frac{1}{4}\khat + \mathbf{D} \\ \\ \mathbf{D} &= \jhat + \frac{1}{4}\khat \end{align*} $$
Therefore, the position vector is

$$ \mathbf{r}(t) = \left(\frac{t^3}{3} + t\right)\,\ihat + \left(t - \sin t + 1\right)\,\jhat + \left(\frac{1}{4} - \frac{1}{4}\cos 2t\right)\,\khat $$

13.4.3 Projectile Motion

A projectile is an object that is in free fall (acting only under the influence of gravity) after some initial velocity is applied to it. Projectiles are a common application of kinematics and a problem of interest in beginning physics courses. Often, these problems are presented with a body such as a ball being given an initial velocity and an angle to the ground in which that velocity is applied. You must then use trigonometry to resolve the components of the initial velocity to reconstruct the position function. Also, questions about the system such as "how far from the initial point on the ground did the object travel" or "what was the highest altitude achieved by the object" are considered.

The procedure for tackling these problems is the same, determine the vector equation describing the position function and use it alongside some algebra and calculus to answer any questions about the system.

Example 13.4.2

A projectile is fired with an initial speed of $200\text{m/s}$ and and angle of elevation $60^\circ$. Find

1. the range of the projectile
2. the maximum height reached
3. the speed at impact

Solution

Since gravity is the only force in play, we can treat this problem as only having two dimensions of interest: the horizontal direction and the vertical direction. Assuming the terrain is flat, the range of the projectile is the horizontal distance traveled before it hits the ground. We will let this be in the $\ihat$ direction, and the vertical position be in the $\jhat$ direction. For simplicity, we will also assume that the projectile was fired from close enough to the ground that the height of the tank is negligible.

In mathematical modeling, when we assume something is zero in a problem, it is not because it "doesn't exist," but because we are making a reasonable assumption that quantity will not greatly affect the final result.

Before finding the range of the projectile, we have to find the vertical and horizontal components of the initial velocity. We do this by using trigonometry and knowing that at and angle of $60^\circ$, the horizontal component is given by cosine and the vertical component by sine. Hence,


$$ \begin{align*} \mathbf{v}(0) &= \left(200\cos (60^\circ)\right)\,\ihat + \left(200\sin (60^\circ)\right)\,\jhat \\ \\ &= 100\,\ihat + 100\sqrt{3}\,\jhat \end{align*} $$
We now will start with $\mathbf{a}(t) = -g\,\jhat \approx \left(9.8\text{ m/s^2}\right)\,\jhat$ and integrate until we have the position function.

$$ \begin{align*} \mathbf{v}(t) &= \int \mathbf{a}(t)\,dt \\ \\ &= 0\,\ihat - 9.8t\,\jhat + \mathbf{C} \end{align*} $$
Applying the initial velocity:

$$ \begin{align*} \mathbf{v}(0) &= 100\,\ihat + 100\sqrt{3}\,\jhat = 0\,\ihat - 9.8(0)\,\jhat + \mathbf{C} \\ \\ \mathbf{C} &= 100\,\ihat + 100\sqrt{3}\,\jhat \end{align*} $$
So the velocity function is

$$ \mathbf{v}(t) = 100\,\ihat + \left(100\sqrt{3} - 9.8t\right)\,\jhat $$
We integrate again to find $\mathbf{r}(t)$

$$ \mathbf{r}(t) = 100t\,\ihat + \left(100\sqrt{3}t - 4.9t^2\right)\,\jhat $$
The initial position is $\mathbf{0}$, so it is pretty obvious that the constant of integration will be $\mathbf{0}$. Therefore the above $\mathbf{r}(t)$ is the position function for our tank shell.

1. Projectile Range
   To find the range of the projectile, we need to find the times when the vertical component of the position vector is zero. We have the function
   
   $$ r_y(t) = 100\sqrt{3}t - 4.9t^2 = t\left(100\sqrt{3} - 4.9t\right) $$
   whose roots are $t = 0$ and $t = \frac{100\sqrt{3}}{4.9}\approx 35.3\text{ s}$. The second value is our answer for where the projectile strikes the ground, and so we place this value of $t$ into the horizontal component equation $r_x(t) = 100t$ to find that the particle's range is approximately $3535\text{ m}$.

2. Maximum Height
   Maximum height is achieved when the vertical component of the displacement has a zero derivative, or equivalently, when the velocity is zero. We know that the vertical component of the velocity function is
   
   $$ v_y(t) = 100\sqrt{3} - 9.8t $$
   and so this is zero when $t = \frac{100\sqrt{3}}{9.8}\approx 17.7\text{ s}$. We plug this value into the vertical position function to find that the maximum height is
   
   
   $$ r_y\left(\frac{100\sqrt{3}}{9.8}\right) = 100\sqrt{3}\left(\frac{100\sqrt{3}}{9.8}\right) - 4.9\left(\frac{100\sqrt{3}}{9.8}\right)^2 \approx 1531\text{ m}$$
   

3. Speed at Impact
   We already know that the impact occurs at $t = \frac{100\sqrt{3}}{4.9}$, so this is plugged into the velocity function and then we find the magnitude of the vector.
   
   $$ \begin{align*} \mathbf{v}\left(\frac{100\sqrt{3}}{4.9}\right) &= 100\,\ihat + \left(100\sqrt{3} - 9.8\left(\frac{100\sqrt{3}}{4.9}\right)\right)\,\jhat \\ \\ &= 100\,\ihat - 100\sqrt{3}\,\jhat \end{align*} $$ The speed at impact is given by
   
   $$ \left|\mathbf{v}'(t)\right| = \sqrt{(100)^2 + (-100\sqrt{3})^2} \approx 200\text{ m/s} $$
   Here is a sketch of the trajectory.

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