The Central Idea
For motion along a curved path, acceleration can be decomposed into two perpendicular directions:
- the tangential direction, which follows the path;
- the normal direction, which points toward the local center of curvature.
The acceleration vector can therefore be written
\[ \boxed{ \mathbf{a} = a_T\mathbf{T} + a_N\mathbf{N} }. \]Why Decompose Acceleration?
The ordinary coordinate components
\[ \langle a_x,a_y\rangle \]describe acceleration relative to fixed axes.
Tangential and normal components instead describe acceleration relative to the moving path itself.
This makes it easier to answer two questions:
- How quickly is the speed changing?
- How quickly is the direction changing?
The Unit Tangent Vector
Let
\[ \mathbf{r}(t) \]be a position vector and
\[ \mathbf{v}(t) = \mathbf{r}'(t) \]its velocity.
When
\[ \mathbf{v}(t)\ne\mathbf{0}, \]the unit tangent vector is
\[ \boxed{ \mathbf{T}(t) = \frac{\mathbf{v}(t)} {|\mathbf{v}(t)|} }. \]What the Tangent Vector Tells Us
The vector
\[ \mathbf{T} \]has magnitude \(1\) and points in the instantaneous direction of motion.
Velocity can therefore be written
\[ \boxed{ \mathbf{v} = v\mathbf{T} } \]where
\[ v=|\mathbf{v}| \]is the scalar speed.
Speed and Direction Are Separated
The expression
\[ \mathbf{v}=v\mathbf{T} \]separates velocity into:
- a magnitude \(v\), which tells how fast the object moves;
- a unit direction \(\mathbf{T}\), which tells where it is headed.
Differentiating this product reveals the two parts of acceleration.
Differentiate the Velocity
Starting from
\[ \mathbf{v}=v\mathbf{T}, \]differentiate with respect to time:
\[ \mathbf{a} = \frac{d}{dt} \left( v\mathbf{T} \right). \]By the product rule,
\[ \boxed{ \mathbf{a} = \frac{dv}{dt}\mathbf{T} + v\frac{d\mathbf{T}}{dt} }. \]The Tangential Component
The first term,
\[ \frac{dv}{dt}\mathbf{T}, \]lies along the tangent direction.
Therefore, the tangential acceleration component is
\[ \boxed{ a_T = \frac{dv}{dt} }. \]It measures the rate at which speed changes.
Interpreting \(a_T\)
If
\[ a_T>0, \]speed is increasing.
If
\[ a_T<0, \]speed is decreasing.
If
\[ a_T=0, \]speed is momentarily constant.
The Normal Direction
The unit tangent vector changes direction as the path curves.
The derivative
\[ \mathbf{T}'(t) \]points toward the side on which the tangent direction is turning.
When
\[ \mathbf{T}'(t)\ne\mathbf{0}, \]define the unit normal vector by
\[ \boxed{ \mathbf{N}(t) = \frac{\mathbf{T}'(t)} {|\mathbf{T}'(t)|} }. \]The Tangent and Normal Vectors Are Perpendicular
Since
\[ |\mathbf{T}|=1, \]we have
\[ \mathbf{T}\cdot\mathbf{T}=1. \]Differentiate:
\[ 2\mathbf{T}\cdot\mathbf{T}'=0. \]Therefore,
\[ \mathbf{T}\cdot\mathbf{T}'=0. \]Since \(\mathbf{N}\) points in the direction of \(\mathbf{T}'\),
\[ \boxed{ \mathbf{T}\perp\mathbf{N} }. \]The Normal Component of Acceleration
The second part of acceleration,
\[ v\frac{d\mathbf{T}}{dt}, \]is perpendicular to the tangent.
It measures how the direction of motion changes.
This is the normal component of acceleration.
Curvature
Curvature measures how rapidly the tangent direction changes with distance traveled along the path.
If \(s\) denotes arc length, curvature is
\[ \boxed{ \kappa = \left| \frac{d\mathbf{T}}{ds} \right| }. \]Large curvature means the path bends sharply.
Small curvature means the path bends gently.
Radius of Curvature
The radius of curvature \(\rho\) is the reciprocal of curvature:
\[ \boxed{ \rho=\frac1\kappa }. \]A large radius of curvature corresponds to a gently bending path.
A small radius of curvature corresponds to a sharply bending path.
Relating Time and Arc Length
Speed is the rate at which arc length changes:
\[ \boxed{ \frac{ds}{dt}=v }. \]Therefore,
\[ \frac{d\mathbf{T}}{dt} = \frac{d\mathbf{T}}{ds} \frac{ds}{dt}. \]Taking magnitudes,
\[ \left| \frac{d\mathbf{T}}{dt} \right| = \kappa v. \]The Normal Acceleration Formula
The magnitude of the normal part of acceleration is
\[ a_N = v \left| \frac{d\mathbf{T}}{dt} \right|. \]Since
\[ \left| \frac{d\mathbf{T}}{dt} \right| = \kappa v, \]we obtain
\[ \boxed{ a_N=\kappa v^2 }. \]Normal Acceleration and Radius of Curvature
Since
\[ \kappa=\frac1\rho, \]the normal acceleration can also be written
\[ \boxed{ a_N=\frac{v^2}{\rho} }. \]This is the general curved-path version of the centripetal acceleration formula.
The Full Tangential-Normal Decomposition
Combining the two components gives
\[ \boxed{ \mathbf{a} = a_T\mathbf{T} + a_N\mathbf{N} }. \]With the component formulas,
\[ \boxed{ \mathbf{a} = \frac{dv}{dt}\mathbf{T} + \frac{v^2}{\rho}\mathbf{N} }. \]What Each Component Does
| Component | Formula | Main Effect |
|---|---|---|
| Tangential | \(a_T=dv/dt\) | Changes speed |
| Normal | \(a_N=v^2/\rho\) | Changes direction |
Uniform Circular Motion as a Special Case
For uniform circular motion,
\[ v=\text{constant}. \]Therefore,
\[ a_T=\frac{dv}{dt}=0. \]The radius of curvature is the circle radius:
\[ \rho=r. \]Thus
\[ \boxed{ a_N=\frac{v^2}{r} }. \]Why Circular Acceleration Is Purely Normal
In uniform circular motion, speed does not change.
Therefore, acceleration cannot have a tangential component.
The entire acceleration is devoted to turning the velocity vector:
\[ \boxed{ \mathbf{a} = \frac{v^2}{r}\mathbf{N} }. \]Straight-Line Motion as Another Special Case
A straight line has zero curvature:
\[ \kappa=0. \]Equivalently,
\[ \rho=\infty \]in the limiting geometric sense.
Therefore,
\[ \boxed{ a_N=0 }. \]Any acceleration must be purely tangential.
Changing Speed on a Straight Line
If an object moves along a straight line and speeds up, then
\[ a_T>0 \]while
\[ a_N=0. \]If it slows down,
\[ a_T<0 \]while the normal component remains zero.
Constant Speed on a Curved Path
An object can move at constant speed along a noncircular curved path.
Then
\[ a_T=0, \]but generally
\[ a_N\ne0. \]The size of the normal acceleration changes as the local curvature changes.
Changing Speed and Direction Together
In general curved motion, both components can be nonzero:
\[ a_T\ne0 \]and
\[ a_N\ne0. \]Then the object is simultaneously changing speed and turning.
Magnitude of the Acceleration Vector
Since
\[ \mathbf{T}\perp\mathbf{N}, \]the tangential and normal components form a right triangle.
Therefore,
\[ \boxed{ |\mathbf{a}| = \sqrt{ a_T^2+a_N^2 } }. \]Finding the Tangential Component from Velocity and Acceleration
Since
\[ \mathbf{T} = \frac{\mathbf{v}}{|\mathbf{v}|}, \]the tangential component is
\[ a_T = \mathbf{a}\cdot\mathbf{T}. \]Therefore,
\[ \boxed{ a_T = \frac{ \mathbf{v}\cdot\mathbf{a} }{ |\mathbf{v}| } }. \]Why the Dot-Product Formula Works
Begin with
\[ \mathbf{a} = a_T\mathbf{T} + a_N\mathbf{N}. \]Dot both sides with \(\mathbf{T}\):
\[ \mathbf{a}\cdot\mathbf{T} = a_T(\mathbf{T}\cdot\mathbf{T}) + a_N(\mathbf{N}\cdot\mathbf{T}). \]Since
\[ \mathbf{T}\cdot\mathbf{T}=1 \]and
\[ \mathbf{N}\cdot\mathbf{T}=0, \]we get
\[ \boxed{ \mathbf{a}\cdot\mathbf{T}=a_T }. \]Finding the Normal Component from Magnitudes
If \(|\mathbf{a}|\) and \(a_T\) are known, then
\[ |\mathbf{a}|^2 = a_T^2+a_N^2. \]Therefore,
\[ \boxed{ a_N = \sqrt{ |\mathbf{a}|^2-a_T^2 } }. \]By convention, \(a_N\ge0\).
Example: Finding \(a_T\) and \(a_N\) from Vectors
Suppose
\[ \mathbf{v} = \langle3,4\rangle \]and
\[ \mathbf{a} = \langle2,1\rangle. \]First,
\[ |\mathbf{v}|=5. \]Then
\[ a_T = \frac{ \mathbf{v}\cdot\mathbf{a} }{ |\mathbf{v}| }. \]Compute the dot product:
\[ \mathbf{v}\cdot\mathbf{a} = 3(2)+4(1)=10. \]Therefore,
\[ \boxed{ a_T=2 }. \]Continuing the Same Example
The acceleration magnitude is
\[ |\mathbf{a}| = \sqrt{2^2+1^2} \] \[ = \sqrt5. \]Therefore,
\[ a_N = \sqrt{ 5-2^2 } \] \[ = \boxed{1}. \]So the acceleration has tangential magnitude \(2\) and normal magnitude \(1\).
The Tangent Vector in the Same Example
Since
\[ \mathbf{v} = \langle3,4\rangle \]and
\[ |\mathbf{v}|=5, \]the unit tangent vector is
\[ \boxed{ \mathbf{T} = \left\langle \frac35, \frac45 \right\rangle }. \]Finding the Normal Direction
Once \(a_T\mathbf{T}\) is known, subtract it from the full acceleration:
\[ \mathbf{a}_N = \mathbf{a}-a_T\mathbf{T}. \]If \(a_N\ne0\), then
\[ \boxed{ \mathbf{N} = \frac{ \mathbf{a}-a_T\mathbf{T} }{ a_N } }. \]Example: Recovering the Normal Vector
Continue with
\[ \mathbf{a} = \langle2,1\rangle, \] \[ a_T=2, \]and
\[ \mathbf{T} = \left\langle \frac35, \frac45 \right\rangle. \]Then
\[ a_T\mathbf{T} = \left\langle \frac65, \frac85 \right\rangle. \]Therefore,
\[ \mathbf{a}_N = \left\langle 2-\frac65, 1-\frac85 \right\rangle \] \[ = \left\langle \frac45, -\frac35 \right\rangle. \]Since \(a_N=1\),
\[ \boxed{ \mathbf{N} = \left\langle \frac45, -\frac35 \right\rangle }. \]Speeding Up and Slowing Down
The sign of
\[ a_T \]directly describes speed change:
\[ \boxed{ a_T>0 \Rightarrow \text{speed increasing} } \] \[ \boxed{ a_T<0 \Rightarrow \text{speed decreasing} }. \]This avoids having to compare coordinate signs separately.
Relation to \(\mathbf{v}\cdot\mathbf{a}\)
Since
\[ a_T = \frac{ \mathbf{v}\cdot\mathbf{a} }{ |\mathbf{v}| }, \]the sign of
\[ \mathbf{v}\cdot\mathbf{a} \]also reveals speed change.
Thus:
\[ \boxed{ \mathbf{v}\cdot\mathbf{a}>0 \Rightarrow \text{speed increasing} } \] \[ \boxed{ \mathbf{v}\cdot\mathbf{a}<0 \Rightarrow \text{speed decreasing} }. \]When \(\mathbf{v}\cdot\mathbf{a}=0\)
If
\[ \mathbf{v}\cdot\mathbf{a}=0, \]then
\[ a_T=0. \]The speed is momentarily constant.
The acceleration may still be nonzero because the normal component can remain.
Uniform Circular Motion Revisited
In uniform circular motion,
\[ \mathbf{v}\cdot\mathbf{a}=0 \]at every instant.
Therefore,
\[ a_T=0 \]and all acceleration is normal.
Curvature of a Plane Parametric Curve
For
\[ \mathbf{r}(t) = \langle x(t),y(t)\rangle, \]a useful curvature formula is
\[ \boxed{ \kappa = \frac{ |x'y''-y'x''| }{ \left( x'^2+y'^2 \right)^{3/2} } }. \]This formula connects coordinate derivatives directly to geometric bending.
Radius of Curvature from Coordinate Derivatives
Once curvature is known,
\[ \rho=\frac1\kappa. \]Therefore,
\[ \boxed{ \rho = \frac{ \left( x'^2+y'^2 \right)^{3/2} }{ |x'y''-y'x''| } } \]whenever the denominator is nonzero.
Example: Curvature of a Parabolic Motion
Let
\[ \mathbf{r}(t) = \langle t,t^2\rangle. \]Then
\[ x'=1, \qquad y'=2t, \]and
\[ x''=0, \qquad y''=2. \]Therefore,
\[ \kappa = \frac{ |1(2)-(2t)(0)| }{ (1+4t^2)^{3/2} }. \]Thus
\[ \boxed{ \kappa = \frac{ 2 }{ (1+4t^2)^{3/2} } }. \]Speed of the Parabolic Motion
For
\[ \mathbf{r}(t) = \langle t,t^2\rangle, \]velocity is
\[ \mathbf{v}(t) = \langle1,2t\rangle. \]Therefore,
\[ v = \sqrt{1+4t^2}. \]Normal Acceleration of the Parabolic Motion
Use
\[ a_N=\kappa v^2. \]Then
\[ a_N = \frac{2}{(1+4t^2)^{3/2}} (1+4t^2). \]Therefore,
\[ \boxed{ a_N = \frac{ 2 }{ \sqrt{1+4t^2} } }. \]Tangential Acceleration of the Parabolic Motion
Since
\[ v = \sqrt{1+4t^2}, \]differentiate:
\[ a_T = \frac{dv}{dt}. \]Therefore,
\[ \boxed{ a_T = \frac{ 4t }{ \sqrt{1+4t^2} } }. \]Checking the Parabolic Example from the Acceleration Vector
For
\[ \mathbf{r}(t) = \langle t,t^2\rangle, \]acceleration is
\[ \mathbf{a} = \langle0,2\rangle. \]Therefore,
\[ |\mathbf{a}|=2. \]The decomposition must satisfy
\[ a_T^2+a_N^2=4. \]Substituting the formulas confirms this identity.
Curvature in Three Dimensions
For a sufficiently smooth space curve,
\[ \mathbf{r}(t), \]curvature may be computed from
\[ \boxed{ \kappa = \frac{ |\mathbf{v}\times\mathbf{a}| }{ |\mathbf{v}|^3 } }. \]This formula is especially useful when a three-dimensional vector path is already given.
Normal Acceleration in Three Dimensions
The same structural formula holds:
\[ \boxed{ a_N=\kappa v^2 }. \]Using the cross-product curvature formula,
\[ a_N = \frac{ |\mathbf{v}\times\mathbf{a}| }{ |\mathbf{v}| }. \]Therefore,
\[ \boxed{ a_N = \frac{ |\mathbf{v}\times\mathbf{a}| }{ |\mathbf{v}| } }. \]Example: Circular Motion from Tangential-Normal Form
Suppose an object moves around a circle of radius
\[ r=10\text{ m} \]at constant speed
\[ v=15\frac{\text{m}}{\text{s}}. \]Since speed is constant,
\[ a_T=0. \]The normal component is
\[ a_N = \frac{15^2}{10} \] \[ = \boxed{ 22.5\frac{\text{m}}{\text{s}^2} }. \]Example: Curved Motion with Changing Speed
Suppose at one instant an object has:
\[ v=8\frac{\text{m}}{\text{s}}, \] \[ \frac{dv}{dt} = 3\frac{\text{m}}{\text{s}^2}, \]and radius of curvature
\[ \rho=16\text{ m}. \]Then
\[ a_T=3 \]and
\[ a_N = \frac{8^2}{16} \] \[ = 4. \]Therefore,
\[ |\mathbf{a}| = \sqrt{3^2+4^2} \] \[ = \boxed{ 5\frac{\text{m}}{\text{s}^2} }. \]The Direction of Total Acceleration
The acceleration vector is the vector sum of its tangential and normal parts.
If both components are nonzero, the total acceleration points somewhere between the tangent and normal directions.
Its exact direction depends on the relative sizes of
\[ a_T \]and
\[ a_N. \]Angle of the Acceleration Relative to the Tangent
If \(\phi\) is the angle from the positive tangent direction toward the normal direction, then
\[ \tan\phi = \frac{a_N}{a_T} \]when \(a_T\ne0\).
Thus
\[ \boxed{ \phi = \tan^{-1} \left( \frac{a_N}{a_T} \right) } \]with quadrant interpretation as needed.
Normal Acceleration Is Never Negative
The scalar component
\[ a_N \]is conventionally taken as nonnegative.
The direction is carried by the unit normal vector
\[ \mathbf{N}. \]Therefore,
\[ \boxed{ a_N\ge0 }. \]Tangential Acceleration Can Be Positive or Negative
Unlike the normal component,
\[ a_T \]is signed.
Its sign indicates whether speed is increasing or decreasing along the direction of motion.
When the Normal Vector Is Undefined
If the path is locally straight, then
\[ \mathbf{T}'=\mathbf{0}. \]The principal normal direction is then not determined by curvature at that instant.
In such a case,
\[ a_N=0. \]When the Tangent Vector Is Undefined
The usual formula
\[ \mathbf{T} = \frac{\mathbf{v}}{|\mathbf{v}|} \]requires
\[ \mathbf{v}\ne\mathbf{0}. \]At an instant of zero velocity, the standard tangential-normal decomposition may require a separate limiting analysis.
Connection to Projectile Motion
A projectile has constant downward acceleration:
\[ \mathbf{a} = \langle0,-g\rangle. \]But its tangential and normal components change during flight because the tangent direction changes.
The same fixed acceleration vector can therefore produce different path-relative components at different times.
At the Top of a Projectile
At the top of a nonvertical projectile trajectory, velocity is horizontal.
Gravity points vertically downward.
Therefore,
\[ \mathbf{v}\perp\mathbf{a}. \]So
\[ \boxed{ a_T=0 } \]at that instant, while all acceleration is normal.
Before the Top of a Projectile
During upward flight, gravity has a component opposite the direction of velocity.
Therefore,
\[ a_T<0. \]The projectile's speed decreases as it rises.
After the Top of a Projectile
During downward flight, gravity has a component in the direction of velocity.
Therefore,
\[ a_T>0. \]The projectile's speed increases as it falls.
Common Error: Thinking Tangential Acceleration Means Tangential Velocity
Velocity is always tangent to a smooth path.
Tangential acceleration is only the part of acceleration that lies along the tangent.
The two quantities have different meanings and different units.
Common Error: Thinking Normal Acceleration Means the Object Moves Inward
Normal acceleration changes the direction of the velocity vector.
It does not mean the object leaves its path and moves directly inward.
The instantaneous velocity remains tangent to the path.
Common Error: Using \(a_N=v^2/r\) with the Wrong Radius
For a general curved path, the denominator is the local radius of curvature:
\[ \boxed{ a_N=\frac{v^2}{\rho} }. \]Only for circular motion does
\[ \rho=r \]equal the fixed circle radius.
Common Error: Assuming \(a_T=0\) Means \(\mathbf{a}=0\)
If
\[ a_T=0, \]speed is not changing.
But the normal component may still be nonzero:
\[ a_N>0. \]Uniform circular motion is the standard example.
Common Error: Assuming \(a_N=0\) Means No Acceleration
If the path is straight,
\[ a_N=0, \]but the object can still speed up or slow down.
Then acceleration is purely tangential.
Common Error: Forgetting That \(a_T\) Is Signed
The normal component is conventionally nonnegative.
The tangential component can be negative.
A negative
\[ a_T \]means speed is decreasing.
Common Error: Using \(|\mathbf{a}|\) as the Normal Component
In general,
\[ |\mathbf{a}| \ne a_N. \]They are equal only when
\[ a_T=0. \]The correct magnitude relation is
\[ \boxed{ |\mathbf{a}|^2 = a_T^2+a_N^2 }. \]A Reliable Procedure from \(\mathbf{r}(t)\)
- Differentiate \(\mathbf{r}(t)\) to find \(\mathbf{v}(t)\).
- Compute speed \(v=|\mathbf{v}|\).
- Form \(\mathbf{T}=\mathbf{v}/|\mathbf{v}|\).
- Differentiate speed to find \(a_T=dv/dt\).
- Differentiate again to find \(\mathbf{a}(t)\).
- Use \(a_N=\sqrt{|\mathbf{a}|^2-a_T^2}\).
- Find \(\mathbf{N}\) if a normal direction is required.
A Reliable Procedure from \(\mathbf{v}\) and \(\mathbf{a}\)
- Compute \(v=|\mathbf{v}|\).
- Compute \(\mathbf{T}=\mathbf{v}/v\).
- Find \(a_T=(\mathbf{v}\cdot\mathbf{a})/v\).
- Find \(a_N=\sqrt{|\mathbf{a}|^2-a_T^2}\).
- Compute \(\mathbf{N}=(\mathbf{a}-a_T\mathbf{T})/a_N\) when \(a_N\ne0\).
- Check that \(\mathbf{T}\cdot\mathbf{N}=0\).
A Reliable Procedure When Curvature Is Known
- Identify speed \(v\).
- Differentiate speed to obtain \(a_T\).
- Use \(a_N=\kappa v^2\) or \(a_N=v^2/\rho\).
- Combine components as \(\mathbf{a}=a_T\mathbf{T}+a_N\mathbf{N}\).
- Use the Pythagorean relation for the total acceleration magnitude.
Example: Full Tangential-Normal Workflow
Let
\[ \mathbf{r}(t) = \langle t,t^2\rangle. \]Velocity:
\[ \mathbf{v}(t) = \langle1,2t\rangle. \]Speed:
\[ v = \sqrt{1+4t^2}. \]Tangential component:
\[ a_T = \frac{dv}{dt} \] \[ = \boxed{ \frac{ 4t }{ \sqrt{1+4t^2} } }. \]Acceleration:
\[ \mathbf{a} = \langle0,2\rangle, \]so
\[ |\mathbf{a}|=2. \]Therefore,
\[ a_N = \sqrt{ 4- \frac{ 16t^2 }{ 1+4t^2 } }. \]Simplifying,
\[ \boxed{ a_N = \frac{ 2 }{ \sqrt{1+4t^2} } }. \]Practice 1: Unit Tangent Vector
Suppose
\[ \mathbf{v} = \langle6,8\rangle. \]Find \(\mathbf{T}\).
Practice 2: Tangential Component
An object's speed is
\[ v(t)=t^2+3. \]Find \(a_T(t)\).
Practice 3: Normal Component from Radius of Curvature
An object moves at
\[ 12\frac{\text{m}}{\text{s}} \]on a path whose radius of curvature is
\[ 18\text{ m}. \]Find \(a_N\).
Practice 4: Uniform Circular Motion
An object moves at constant speed
\[ 10\frac{\text{m}}{\text{s}} \]on a circle of radius
\[ 5\text{ m}. \]Find \(a_T\) and \(a_N\).
Practice 5: Acceleration Magnitude
Suppose
\[ a_T=6 \]and
\[ a_N=8. \]Find \(|\mathbf{a}|\).
Practice 6: Tangential Component from Vectors
Let
\[ \mathbf{v} = \langle4,3\rangle \]and
\[ \mathbf{a} = \langle2,1\rangle. \]Find \(a_T\).
Practice 7: Normal Component from Vectors
Continue Practice 6 and find \(a_N\).
Practice 8: Curvature
Let
\[ \mathbf{r}(t) = \langle t,t^2\rangle. \]Find the curvature at \(t=0\).
Practice 9: Speed Change
Suppose
\[ \mathbf{v}\cdot\mathbf{a}<0. \]Is the object speeding up or slowing down?
Practice 10: Identify the Error
A student says:
“If the tangential acceleration is zero, then the total acceleration must also be zero.”
Explain the error.
Practice Answers
Practice 1
\[ |\mathbf{v}| = \sqrt{6^2+8^2} \] \[ =10. \]Therefore,
\[ \boxed{ \mathbf{T} = \left\langle \frac35, \frac45 \right\rangle }. \]Practice 2
\[ a_T = \frac{dv}{dt} \] \[ = \boxed{2t}. \]Practice 3
\[ a_N = \frac{v^2}{\rho} \] \[ = \frac{12^2}{18} \] \[ = \frac{144}{18} \] \[ = \boxed{ 8\frac{\text{m}}{\text{s}^2} }. \]Practice 4
Constant speed gives
\[ \boxed{ a_T=0 }. \]Normal acceleration:
\[ a_N = \frac{10^2}{5} \] \[ = \boxed{ 20\frac{\text{m}}{\text{s}^2} }. \]Practice 5
\[ |\mathbf{a}| = \sqrt{6^2+8^2} \] \[ = \boxed{10}. \]Practice 6
\[ |\mathbf{v}| = \sqrt{4^2+3^2} \] \[ =5. \]The dot product is
\[ \mathbf{v}\cdot\mathbf{a} = 4(2)+3(1) \] \[ =11. \]Therefore,
\[ \boxed{ a_T = \frac{11}{5} }. \]Practice 7
\[ |\mathbf{a}| = \sqrt{2^2+1^2} \] \[ = \sqrt5. \]Therefore,
\[ a_N = \sqrt{ 5- \left( \frac{11}{5} \right)^2 }. \] \[ = \sqrt{ \frac{125-121}{25} }. \] \[ = \boxed{ \frac25 }. \]Practice 8
For
\[ x=t, \qquad y=t^2, \]we have
\[ x'=1, \qquad y'=2t, \qquad x''=0, \qquad y''=2. \]Thus
\[ \kappa = \frac{ 2 }{ (1+4t^2)^{3/2} }. \]At \(t=0\),
\[ \boxed{\kappa=2}. \]Practice 9
Since
\[ a_T = \frac{ \mathbf{v}\cdot\mathbf{a} }{ |\mathbf{v}| }, \]a negative dot product gives
\[ a_T<0. \]Therefore, the object is slowing down.
Practice 10
Zero tangential acceleration means only that speed is not changing.
The normal component may still be nonzero:
\[ a_N>0. \]In uniform circular motion, for example,
\[ a_T=0 \]while
\[ \boxed{ a_N=\frac{v^2}{r} }. \]The object accelerates because its direction changes.
What to Remember
The unit tangent vector is
\[ \boxed{ \mathbf{T} = \frac{\mathbf{v}}{|\mathbf{v}|} }. \]The unit normal vector is
\[ \boxed{ \mathbf{N} = \frac{\mathbf{T}'} {|\mathbf{T}'|} } \]when \(\mathbf{T}'\ne\mathbf{0}\).
Acceleration decomposes as
\[ \boxed{ \mathbf{a} = a_T\mathbf{T} + a_N\mathbf{N} }. \]The component formulas are
\[ \boxed{ a_T = \frac{dv}{dt} } \]and
\[ \boxed{ a_N = \kappa v^2 = \frac{v^2}{\rho} }. \]Also,
\[ \boxed{ a_T = \frac{ \mathbf{v}\cdot\mathbf{a} }{ |\mathbf{v}| } } \]and
\[ \boxed{ |\mathbf{a}| = \sqrt{ a_T^2+a_N^2 } }. \]The main conceptual distinctions are:
- tangential acceleration changes speed;
- normal acceleration changes direction;
- constant speed can still produce nonzero normal acceleration;
- straight-line motion has zero normal acceleration;
- curvature controls how strongly a path turns;
- the radius of curvature replaces the fixed circle radius for general curves.
Where We Go Next
Tangential and normal components organize acceleration geometrically along a path. The final lesson in this section returns to graphical representations and connects the shapes, slopes, signs, and areas of position, velocity, and acceleration graphs.
The next lesson, Position, Velocity, and Acceleration Graphs, synthesizes the main kinematic relationships developed throughout this section.