Velocity And Acceleration For Polynomial Position Models

From Position Functions to Motion

A polynomial position model describes an object's location at each time by giving polynomial functions for its coordinates. Differentiating those coordinates gives a direct way to calculate the object's velocity and acceleration at any time in the model's domain.

The calculations are coordinatewise, but the quantities have different meanings: position records location, velocity is a vector describing change in position per unit time, and acceleration is a vector describing change in velocity per unit time. Speed is a scalar: the magnitude of the velocity vector.

Key idea: Differentiate each position coordinate once to obtain velocity, and differentiate each velocity coordinate once more to obtain acceleration. Evaluate the resulting functions at the requested time.

Polynomial Position, Velocity, and Acceleration

For motion in $n$-dimensional space, write the position as a vector-valued function of time:

\mathbf r(t)=\langle x_1(t),x_2(t),\ldots,x_n(t)\rangle.

Here each coordinate $x_i(t)$ gives the object's position along one coordinate axis. If each $x_i$ is a polynomial, then each coordinate can be differentiated using the usual polynomial differentiation rules. The velocity vector is the derivative of position, and acceleration is the derivative of velocity.

Definition: For a position function $\mathbf r(t)=\langle x_1(t),\ldots,x_n(t)\rangle$, its velocity and acceleration are $$ \mathbf v(t)=\mathbf r'(t)=\langle x_1'(t),\ldots,x_n'(t)\rangle, \qquad \mathbf a(t)=\mathbf v'(t)=\mathbf r''(t)=\langle x_1''(t),\ldots,x_n''(t)\rangle. $$ The speed at time $t$ is the magnitude of velocity: $$ \|\mathbf v(t)\|=\sqrt{(x_1'(t))^2+\cdots+(x_n'(t))^2}. $$

These expressions apply wherever the position coordinates are defined and twice differentiable. Polynomial coordinates are differentiable for every real $t$, although a physical model may use only a specified time interval. If position is measured in metres and time in seconds, velocity has units of metres per second, while acceleration has units of metres per second squared.

Why Differentiation Works Coordinate by Coordinate

The derivative of a vector-valued function is defined by the limit of its difference quotient, just as for a scalar function. For a polynomial position vector, write the difference quotient by coordinates:

\frac{\mathbf r(t+h)-\mathbf r(t)}{h} = \left\langle \frac{x_1(t+h)-x_1(t)}{h}, \ldots, \frac{x_n(t+h)-x_n(t)}{h} \right\rangle.

As $h$ approaches zero, each coordinate quotient approaches the derivative of that coordinate. A vector in finite-dimensional space converges exactly when each of its coordinates converges to the corresponding coordinate of the limiting vector. Therefore, the limit of the vector difference quotient is the vector of coordinate derivatives. Applying the same argument to $\mathbf v(t)$ shows that acceleration is the vector of second derivatives of the position coordinates.

\mathbf r'(t)=\langle x_1'(t),\ldots,x_n'(t)\rangle, \qquad \mathbf r''(t)=\langle x_1''(t),\ldots,x_n''(t)\rangle.

This gives a useful calculation sequence: first differentiate each coordinate, then substitute the requested time into the position, velocity, and acceleration functions. Compute speed from the velocity at that time, not from the position or acceleration.

A Reliable Calculation Process

Record the coordinates.
Write the position function as a vector and note the time interval and units used by the model.

Differentiate once.
Differentiate each coordinate separately to obtain the velocity vector.

Differentiate again.
Differentiate each velocity coordinate to obtain the acceleration vector.

Evaluate and report.
Substitute the given time into each function. Find speed by taking the magnitude of the evaluated velocity vector.

Worked Polynomial Motion Models

Worked Example: A Cubic and Quadratic Position Model

An object's position, measured in metres, is $\mathbf r(t)=\langle t^3-3t^2+2t+1,\;2t^2-4t\rangle$, where $t$ is measured in seconds and $0\le t\le 3$. Find its position, velocity, acceleration, and speed at $t=2$.

Differentiate the first coordinate and then the second coordinate:

\mathbf v(t)=\mathbf r'(t) =\langle 3t^2-6t+2,\;4t-4\rangle.

Differentiate velocity coordinate by coordinate:

\mathbf a(t)=\mathbf v'(t) =\langle 6t-6,\;4\rangle.

Substitute $t=2$ into the position function:

\mathbf r(2) =\langle 2^3-3(2^2)+2(2)+1,\;2(2^2)-4(2)\rangle =\langle 8-12+4+1,\;8-8\rangle =\langle 1,0\rangle\ \text{m}.

Now evaluate the velocity:

\mathbf v(2) =\langle 3(2^2)-6(2)+2,\;4(2)-4\rangle =\langle 12-12+2,\;8-4\rangle =\langle 2,4\rangle\ \text{m/s}.

The speed is the magnitude of this velocity:

\|\mathbf v(2)\| =\sqrt{2^2+4^2} =\sqrt{4+16} =2\sqrt{5}\ \text{m/s}.

Finally, evaluate acceleration:

\mathbf a(2) =\langle 6(2)-6,\;4\rangle =\langle 12-6,\;4\rangle =\langle 6,4\rangle\ \text{m/s}^2.

Check: Differentiating $t^3-3t^2+2t+1$ gives $3t^2-6t+2$, and differentiating $2t^2-4t$ gives $4t-4$, as used in $\mathbf v(t)$. Differentiating those expressions gives $6t-6$ and $4$, confirming $\mathbf a(t)$. Substitution into the original coordinates gives $\mathbf r(2)=\langle1,0\rangle$, and the velocity components give speed $2\sqrt5\ \text{m/s}$.

Worked Example: A Quadratic Position Model

A position function in metres is $\mathbf r(t)=\langle5+4t-t^2,\;-2+3t+2t^2\rangle$, for $0\le t\le 4$. Find position, velocity, acceleration, and speed at $t=2$.

Differentiate each coordinate:

\mathbf v(t)=\langle 4-2t,\;3+4t\rangle, \qquad \mathbf a(t)=\langle -2,\;4\rangle.

Evaluate position at $t=2$:

\mathbf r(2) =\langle 5+4(2)-2^2,\;-2+3(2)+2(2^2)\rangle =\langle 5+8-4,\;-2+6+8\rangle =\langle 9,12\rangle\ \text{m}.

Evaluate velocity and calculate its magnitude:

\mathbf v(2) =\langle 4-2(2),\;3+4(2)\rangle =\langle 0,11\rangle\ \text{m/s}, \qquad \|\mathbf v(2)\| =\sqrt{0^2+11^2} =11\ \text{m/s}.

Acceleration is constant, so evaluation gives:

\mathbf a(2)=\langle -2,4\rangle\ \text{m/s}^2.

Check: The derivative of $5+4t-t^2$ is $4-2t$, and its second derivative is $-2$. The derivative of $-2+3t+2t^2$ is $3+4t$, and its second derivative is $4$. These reproduce the stated velocity and acceleration. Substitution at $t=2$ gives the reported position and velocity, whose magnitude is $11$.

Worked Example: A Three-Dimensional Model

Consider the position function $\mathbf r(t)=\langle t^2,\;t^3-3t,\;2t^2+1\rangle$, with coordinates in metres and time in seconds. Find the motion quantities at $t=1$.

Differentiate all three coordinates to obtain velocity, and differentiate again to obtain acceleration:

\mathbf v(t)=\langle 2t,\;3t^2-3,\;4t\rangle, \qquad \mathbf a(t)=\langle 2,\;6t,\;4\rangle.

Substitution into the position function gives:

\mathbf r(1) =\langle 1^2,\;1^3-3(1),\;2(1^2)+1\rangle =\langle 1,-2,3\rangle\ \text{m}.

The velocity and speed at $t=1$ are:

\mathbf v(1) =\langle 2(1),\;3(1^2)-3,\;4(1)\rangle =\langle 2,0,4\rangle\ \text{m/s}, $$ $$ \|\mathbf v(1)\| =\sqrt{2^2+0^2+4^2} =\sqrt{4+16} =2\sqrt5\ \text{m/s}.

Acceleration at the same time is:

\mathbf a(1)=\langle 2,6(1),4\rangle =\langle 2,6,4\rangle\ \text{m/s}^2.

Check: The first derivatives of $t^2$, $t^3-3t$, and $2t^2+1$ are respectively $2t$, $3t^2-3$, and $4t$. Their second derivatives are $2$, $6t$, and $4$. Substitution at $t=1$ confirms the listed position, velocity, and acceleration. The squared velocity components sum to $2^2+0^2+4^2=20$, so the speed is $\sqrt{20}=2\sqrt5\ \text{m/s}$.

A Common Differentiation Error

A frequent mistake is to differentiate a power but leave its coefficient unchanged. For example, the derivative of $2t^2$ is not $2t$: the power rule requires multiplying by the exponent and then reducing the exponent by one.

Worked Correction: Differentiating a Coefficient and Power

For $\mathbf r(t)=\langle t^3-2t,\;2t^2\rangle$, a mistaken calculation might use $\mathbf v(t)=\langle 3t^2-2,\;2t\rangle$. At $t=1$, that gives the incorrect velocity $\langle1,2\rangle$ and speed $\sqrt5$.

Apply the power rule correctly to the second coordinate: $\frac{d}{dt}(2t^2)=2(2)t^{2-1}=4t$. Thus the correct velocity and acceleration are:

\mathbf v(t)=\langle 3t^2-2,\;4t\rangle, \qquad \mathbf a(t)=\langle 6t,\;4\rangle.

At $t=1$, these give:

\mathbf v(1)=\langle 3-2,4\rangle=\langle1,4\rangle, \qquad \|\mathbf v(1)\|=\sqrt{1^2+4^2}=\sqrt{17}, \qquad \mathbf a(1)=\langle6,4\rangle.

Check: Differentiating $\langle3t^2-2,4t\rangle$ gives $\langle6t,4\rangle$. In particular, the second velocity coordinate at $t=1$ is $4$, not $2$; the incorrect speed came directly from the incorrect derivative.

Interpreting the Calculated Quantities

At a specified time, position and acceleration are vectors, while speed is a nonnegative number. Keeping those types of quantities distinct prevents common reporting errors. For instance, $\mathbf a(2)=\langle6,4\rangle$ is not a speed, and $\|\mathbf v(2)\|$ is not a velocity vector.

Polynomial models are useful because their derivatives remain simple polynomial functions. Once position is specified, its first and second derivatives provide velocity and acceleration throughout the model's time interval, not only at one chosen instant. Evaluating these functions at a particular time then gives a snapshot of the object's location and its motion quantities at that time.

Takeaway: Differentiate each coordinate separately, substitute the time only after obtaining the relevant function, and calculate speed as the magnitude of velocity. Check each derivative before interpreting the numerical results.

Check Your Understanding

For each question, distinguish the vector quantities from the scalar speed and show the differentiation and substitution steps.

For $\mathbf r(t)=\langle t^2+3t,\;t^3\rangle$, find $\mathbf v(t)$ and $\mathbf a(t)$.
For the same position function, find $\mathbf r(1)$, $\mathbf v(1)$, and $\mathbf a(1)$.
At $t=1$ in Question 2, calculate the speed.
What is the derivative of $5t^3$, and what is its second derivative?
Explain why speed is calculated from velocity rather than from acceleration.