One Motion Problem, Several Connected Quantities
A three-dimensional motion problem may provide acceleration and initial data, or it may give position directly. In either case, the quantities of interest are linked: velocity is the derivative of position, acceleration is the derivative of velocity, and speed is the magnitude of velocity. When velocity is nonzero, acceleration can also be resolved into parts that change the speed and the direction of motion.
The challenge is to keep these relationships connected without confusing vector quantities with scalar quantities. Position, velocity, and acceleration are vectors; speed and the tangential and normal acceleration magnitudes are scalars. Initial conditions select the particular motion from the family of functions obtained by integration.
These formulas apply at a time when the velocity is nonzero. Then the unit tangent vector is \(\mathbf{T}=\mathbf{v}/|\mathbf{v}|\), and the vector tangential component of acceleration is \(a_T\mathbf{T}\). The vector normal component is \(\mathbf{a}-a_T\mathbf{T}\); its magnitude is \(a_N\). If \(\mathbf{v}=\mathbf{0}\), the unit tangent and these formulas are not defined at that instant, so the components must not be calculated by dividing by the speed.
A Reliable Order of Work
When acceleration and initial conditions are given, integrate each coordinate of acceleration to obtain velocity, use the initial velocity to determine the constants, and integrate velocity to obtain position. Then use the initial position to determine the remaining constants. When position is given, differentiate instead. After finding velocity and acceleration, calculate speed and, if required, tangential and normal components. Finally, check the results against the original data and the defining derivative relationships.
Record whether the problem gives position, velocity, or acceleration, along with the relevant initial conditions and time.
Differentiate position when it is given, or integrate acceleration and use initial data when acceleration is given.
Use the magnitude of velocity for speed. Use the dot-product formulas only at times when velocity is nonzero.
Differentiate the proposed position and velocity, and substitute the initial data into the reconstructed functions.
Worked Examples
Worked Example: Reconstructing Motion from Acceleration
A particle moves in three dimensions with acceleration \(\mathbf{a}(t)=\langle 2,6t,-2\rangle\), initial velocity \(\mathbf{v}(0)=\langle 1,0,3\rangle\), and initial position \(\mathbf{r}(0)=\langle 0,2,-1\rangle\). Find its position and velocity, then calculate its speed and tangential and normal acceleration at \(t=1\). Take time in seconds and position in metres.
Integrating acceleration coordinate by coordinate gives \(\mathbf{v}(t)=\langle 2t+C_1,3t^2+C_2,-2t+C_3\rangle\). Substituting \(t=0\) and the given initial velocity gives \(C_1=1\), \(C_2=0\), and \(C_3=3\). Thus \(\mathbf{v}(t)=\langle 2t+1,3t^2,-2t+3\rangle\).
Integrating this velocity gives \(\mathbf{r}(t)=\langle t^2+t+D_1,t^3+D_2,-t^2+3t+D_3\rangle\). The initial position yields \(D_1=0\), \(D_2=2\), and \(D_3=-1\). Therefore,
At \(t=1\), the position is \(\langle 2,3,1\rangle\) metres, the velocity is \(\langle 3,3,1\rangle\) metres per second, and the acceleration is \(\langle 2,6,-2\rangle\) metres per second squared. The speed is \(q=\sqrt{3^2+3^2+1^2}=\sqrt{19}\) metres per second. The dot product is \(\mathbf{v}\cdot\mathbf{a}=3(2)+3(6)+1(-2)=22\), so \(a_T=22/\sqrt{19}\) metres per second squared.
The squared acceleration magnitude is \(2^2+6^2+(-2)^2=44\). Hence \(a_N=\sqrt{44-(22/\sqrt{19})^2}=\sqrt{352/19}=4\sqrt{22}/\sqrt{19}\) metres per second squared. As a vector check, \(\mathbf{T}=\langle 3,3,1\rangle/\sqrt{19}\), so \(a_T\mathbf{T}=\langle 66/19,66/19,22/19\rangle\). Subtracting this from \(\mathbf{a}\) gives the normal component \(\langle -28/19,48/19,-60/19\rangle\), whose squared magnitude is \((784+2304+3600)/361=352/19\), as required.
Finally, differentiating the position gives the stated velocity, and differentiating that velocity gives \(\langle 2,6t,-2\rangle\), the given acceleration. At \(t=0\), the position and velocity are also exactly the prescribed initial values.
Worked Example: Finding Components from a Given Position
Let \(\mathbf{r}(t)=\langle 2\cos t,2\sin t,t^2\rangle\), with position in metres and time in seconds. Find velocity, acceleration, and speed, then determine the tangential and normal acceleration at \(t=0\).
Differentiate each coordinate once to obtain \(\mathbf{v}(t)=\langle -2\sin t,2\cos t,2t\rangle\). Differentiating again gives \(\mathbf{a}(t)=\langle -2\cos t,-2\sin t,2\rangle\). The speed at a general time is \(q(t)=\sqrt{4\sin^2t+4\cos^2t+4t^2}=2\sqrt{1+t^2}\), using \(\sin^2t+\cos^2t=1\).
At \(t=0\), the position is \(\langle 2,0,0\rangle\) metres, the velocity is \(\langle 0,2,0\rangle\) metres per second, and the acceleration is \(\langle -2,0,2\rangle\) metres per second squared. The speed is \(2\), so the velocity is nonzero and the component formulas apply. The dot product is \(0(-2)+2(0)+0(2)=0\); therefore \(a_T=0/2=0\).
The acceleration magnitude is \(\sqrt{(-2)^2+0^2+2^2}=\sqrt{8}=2\sqrt{2}\). Consequently \(a_N=\sqrt{(2\sqrt{2})^2-0^2}=2\sqrt{2}\) metres per second squared. Here \(\mathbf{T}=\langle 0,1,0\rangle\), so the tangential vector component is zero and the normal vector component is the full acceleration \(\langle -2,0,2\rangle\). Its magnitude is indeed \(2\sqrt{2}\).
The derivatives displayed above verify that the calculated velocity and acceleration come from the given position. The nonzero velocity at \(t=0\) also confirms that the component calculation is valid there.
Worked Example: Combining Initial Data with Acceleration Components
Suppose \(\mathbf{a}(t)=\langle 6t,2,-4t\rangle\), \(\mathbf{v}(0)=\langle 0,1,2\rangle\), and \(\mathbf{r}(0)=\langle 1,-2,0\rangle\). Find the motion and the speed, tangential acceleration, and normal acceleration at \(t=1\). Use seconds and metres.
Integrating acceleration gives \(\mathbf{v}(t)=\langle 3t^2+C_1,2t+C_2,-2t^2+C_3\rangle\). The initial velocity sets \(C_1=0\), \(C_2=1\), and \(C_3=2\). Therefore, \(\mathbf{v}(t)=\langle 3t^2,2t+1,-2t^2+2\rangle\). Integrating again gives \(\mathbf{r}(t)=\langle t^3+D_1,t^2+t+D_2,-\frac{2}{3}t^3+2t+D_3\rangle\). The initial position sets \(D_1=1\), \(D_2=-2\), and \(D_3=0\).
At \(t=1\), \(\mathbf{v}=\langle 3,3,0\rangle\) metres per second and \(\mathbf{a}=\langle 6,2,-4\rangle\) metres per second squared. The speed is \(\sqrt{3^2+3^2+0^2}=3\sqrt{2}\) metres per second. The dot product is \(3(6)+3(2)+0(-4)=24\), so \(a_T=24/(3\sqrt{2})=4\sqrt{2}\) metres per second squared.
The acceleration magnitude squared is \(6^2+2^2+(-4)^2=56\). Thus \(a_N=\sqrt{56-(4\sqrt{2})^2}=\sqrt{24}=2\sqrt{6}\) metres per second squared. For a vector check, \(\mathbf{T}=\langle 1/\sqrt{2},1/\sqrt{2},0\rangle\), so the tangential component is \(a_T\mathbf{T}=\langle 4,4,0\rangle\). The remaining component is \(\mathbf{a}-a_T\mathbf{T}=\langle 2,-2,-4\rangle\), whose magnitude is \(\sqrt{2^2+(-2)^2+(-4)^2}=2\sqrt{6}\).
Differentiating the reconstructed position gives \(\langle 3t^2,2t+1,-2t^2+2\rangle\), and differentiating that result gives the original acceleration. Substitution of \(t=0\) into the position and velocity recovers both initial conditions.
Why the Tangential Formula Measures Change in Speed
The tangential component formula follows from the relationship between speed and velocity. At an instant where \(q=|\mathbf{v}|>0\), square the speed to obtain \(q^2=\mathbf{v}\cdot\mathbf{v}\). Differentiating both sides with respect to time gives \(2q\,dq/dt=2\mathbf{v}\cdot\mathbf{a}\). Dividing by \(2q\), which is allowed because \(q>0\), yields
Thus positive \(a_T\) means speed is increasing at that instant, negative \(a_T\) means speed is decreasing, and zero \(a_T\) means the instantaneous rate of change of speed is zero. The normal component is perpendicular to velocity and changes the direction of motion rather than the speed at that instant. Since the tangential and normal components are perpendicular, the Pythagorean theorem gives \(|\mathbf{a}|^2=a_T^2+a_N^2\), which explains the formula for \(a_N\).
A Common Mistake: Dividing by Speed at a Stop
Suppose \(\mathbf{v}(t)=\langle t,0,0\rangle\) and \(\mathbf{a}(t)=\langle 1,0,0\rangle\). At \(t=0\), an incorrect calculation might substitute \(|\mathbf{v}|=0\) into the tangential formula and report \(a_T=(\mathbf{v}\cdot\mathbf{a})/|\mathbf{v}|=0/0=0\). This is not a valid calculation: \(0/0\) is undefined, and the unit tangent \(\mathbf{T}=\mathbf{v}/|\mathbf{v}|\) is also undefined at that instant.
The correction is to state that the tangential-normal formulas based on \(\mathbf{T}\) do not apply at \(t=0\). For \(t>0\), however, \(|\mathbf{v}|=t\) and \(a_T=(t\cdot1)/t=1\). The speed is \(q(t)=t\) for \(t\ge 0\), whose right-hand rate of change at zero is \(1\). That one-sided speed calculation does not make the undefined unit tangent at the instant of rest defined.
Putting the Quantities Together
Integrated motion problems reward careful bookkeeping. Keep vector components in their coordinate order, apply initial conditions to constants of integration, and distinguish speed from velocity. When computing acceleration components, check first that velocity is nonzero, then calculate the dot product and magnitudes, and verify that the tangential and normal parts combine to recover the acceleration.
Check Your Understanding
Use the definitions and checks in this tutorial to answer the following questions.
- If acceleration and initial velocity are known, what information is needed to determine the position function uniquely?
- For \(\mathbf{v}=\langle 1,2,2\rangle\) and \(\mathbf{a}=\langle 2,0,1\rangle\), calculate the speed and tangential acceleration.
- Why must the velocity be nonzero to define the unit tangent vector and use the tangential-normal formulas?
- For \(\mathbf{v}=\langle 2,0,0\rangle\) and \(\mathbf{a}=\langle 0,3,4\rangle\), find the tangential and normal acceleration magnitudes.
- What derivative checks can confirm that position and velocity reconstructed from acceleration are correct?