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Short answer. Tangential acceleration makes the magnitude of the velocity change, while normal acceleration makes the direction of the velocity change.

Some details. The velocity of a point is tangent to its trajectory, and thus we can write the velocity vector as

$\mathbf{v} = v \mathbf{\hat{t}}$,

i.e. as the product of a unit vector $\mathbf{\hat{t}}$, that provides the information about the direction of the vector, and a scalar $v$, the magnitude of the velocity (the number you read on the tachometer of your car, as an example).

Now, we can take time derivative of the velocity, using the law for the derivative of a product (both the magnitude and the direction of the velocity depends on time, in general),

$\dfrac{d \mathbf{v}}{dt} = \dfrac{d}{dt} (v \mathbf{\hat{t}}) = \underbrace{\dfrac{ dv}{dt} \mathbf{\hat{t}}}_{\mathbf{a}_t} + \underbrace{v \dfrac{d \mathbf{\hat{t}}}{dt}}_{\mathbf{a}_n}$,

being the first term tangent to the trajectory (since it has the same direction as $\mathbf{\hat{t}}$), and the second term orthogonal to that, since it's in the direction of the time derivative of a unit vector, that is always orthogonal to the direction itself (or zero, if the direction doesn't change), as we can prove using

$1 = |\mathbf{\hat{t}}|^2 = \mathbf{\hat{t}} \cdot \mathbf{\hat{t}}$$\qquad \rightarrow \qquad$ $0 = \dfrac{d}{dt}|\mathbf{\hat{t}}|^2 = 2 \mathbf{\hat{t}} \cdot \dfrac{d \mathbf{\hat{t}}}{dt}$.

To get a deeper insight into normal acceleration, we need just a bit of differential geometry of curves in space. The time derivative of the unit tangent vector can be written as

$\dfrac{d \mathbf{\hat{t}}}{dt} = k v \mathbf{\hat{n}} = \frac{v}{r} \mathbf{\hat{n}}$$\qquad \rightarrow \qquad$$\mathbf{a}_n = k v^2 \mathbf{\hat{n}} = \dfrac{v^2}{r} \mathbf{\hat{n}}$

using the local unit normal vector $\mathbf{\hat{n}}$ (pointing towards the center of the local circle of curvature) and the local curvature of the trajectory $k$, or the local curvature radius $r = \frac{1}{k}$ (the radius of the local circle of curvature) so that we can recognize the more familiar centripetal acceleration in the normal acceleration. Eventually, we can write the acceleration as

$\dfrac{d \mathbf{v}}{dt} = \mathbf{a}_t + \mathbf{a}_n = \dfrac{ dv}{dt} \mathbf{\hat{t}} + v \dfrac{d \mathbf{\hat{t}}}{dt} = \underbrace{\dfrac{ dv}{dt}}_{a_t} \mathbf{\hat{t}} + \underbrace{k v^2}_{a_n} \mathbf{\hat{n}}$.

Short answer. Tangential acceleration makes the magnitude of the velocity change, while normal acceleration makes the direction of the velocity change.

(I really appreciated the approach of treating kinematics with vectors, and not to rely on a particular choice of a set of coordinates. IMHO it's the right approach)

Some details. The velocity of a point is tangent to its trajectory, and thus we can write the velocity vector as

$\mathbf{v} = v \mathbf{\hat{t}}$,

i.e. as the product of a unit vector $\mathbf{\hat{t}}$, that provides the information about the direction of the vector, and a scalar $v$, the magnitude of the velocity (the number you read on the tachometer of your car, as an example).

Now, we can take time derivative of the velocity, using the law for the derivative of a product (both the magnitude and the direction of the velocity depends on time, in general),

$\dfrac{d \mathbf{v}}{dt} = \dfrac{d}{dt} (v \mathbf{\hat{t}}) = \underbrace{\dfrac{ dv}{dt} \mathbf{\hat{t}}}_{\mathbf{a}_t} + \underbrace{v \dfrac{d \mathbf{\hat{t}}}{dt}}_{\mathbf{a}_n}$,

being the first term tangent to the trajectory (since it has the same direction as $\mathbf{\hat{t}}$), and the second term orthogonal to that, since it's in the direction of the time derivative of a unit vector, that is always orthogonal to the direction itself (or zero, if the direction doesn't change), as we can prove using

$1 = |\mathbf{\hat{t}}|^2 = \mathbf{\hat{t}} \cdot \mathbf{\hat{t}}$$\qquad \rightarrow \qquad$ $0 = \dfrac{d}{dt}|\mathbf{\hat{t}}|^2 = 2 \mathbf{\hat{t}} \cdot \dfrac{d \mathbf{\hat{t}}}{dt}$.

To get a deeper insight into normal acceleration, we need just a bit of differential geometry of curves in space. The time derivative of the unit tangent vector can be written as

$\dfrac{d \mathbf{\hat{t}}}{dt} = k v \mathbf{\hat{n}} = \frac{v}{r} \mathbf{\hat{n}}$$\qquad \rightarrow \qquad$$\mathbf{a}_n = k v^2 \mathbf{\hat{n}} = \dfrac{v^2}{r} \mathbf{\hat{n}}$

using the local unit normal vector $\mathbf{\hat{n}}$ (pointing towards the center of the local circle of curvature) and the local curvature of the trajectory $k$, or the local curvature radius $r = \frac{1}{k}$ (the radius of the local circle of curvature) so that we can recognize the more familiar centripetal acceleration in the normal acceleration. Eventually, we can write the acceleration as

$\dfrac{d \mathbf{v}}{dt} = \mathbf{a}_t + \mathbf{a}_n = \dfrac{ dv}{dt} \mathbf{\hat{t}} + v \dfrac{d \mathbf{\hat{t}}}{dt} = \underbrace{\dfrac{ dv}{dt}}_{a_t} \mathbf{\hat{t}} + \underbrace{k v^2}_{a_n} \mathbf{\hat{n}}$.

Short answer. Tangential acceleration makes the magnitude of the velocity change, while normal acceleration makes the direction of the velocity change.

Some details. The velocity of a point is tangent to its trajectory, and thus we can write the velocity vector as

$\mathbf{v} = v \mathbf{\hat{t}}$,

i.e. as the product of a unit vector $\mathbf{\hat{t}}$, that provides the information about the direction of the vector, and a scalar $v$, the magnitude of the velocity (the number you read on the tachometer of your car, as an example).

Now, we can take time derivative of the velocity, using the law for the derivative of a product (both the magnitude and the direction of the velocity depends on time, in general),

$\dfrac{d \mathbf{v}}{dt} = \dfrac{d}{dt} (v \mathbf{\hat{t}}) = \underbrace{\dfrac{ dv}{dt} \mathbf{\hat{t}}}_{\mathbf{a}_t} + \underbrace{v \dfrac{d \mathbf{\hat{t}}}{dt}}_{\mathbf{a}_n}$,

being the first term tangent to the trajectory (since it has the same direction as $\mathbf{\hat{t}}$), and the second term orthogonal to that, since it's in the direction of the time derivative of a unit vector, that is always orthogonal to the direction itself (or zero, if the direction doesn't change), as we can prove using

$1 = |\mathbf{\hat{t}}|^2 = \mathbf{\hat{t}} \cdot \mathbf{\hat{t}}$$\qquad \rightarrow \qquad$ $0 = \dfrac{d}{dt}|\mathbf{\hat{t}}|^2 = 2 \mathbf{\hat{t}} \cdot \dfrac{d \mathbf{\hat{t}}}{dt}$.

To get a deeper insight into normal acceleration, we need just a bit of differential geometry of curves in space. The time derivative of the unit tangent vector can be written as

$\dfrac{d \mathbf{\hat{t}}}{dt} = k v \mathbf{\hat{n}} = \frac{v}{r} \mathbf{\hat{n}}$$\qquad \rightarrow \qquad$$\mathbf{a}_n = k v^2 \mathbf{\hat{n}} = \dfrac{v^2}{r} \mathbf{\hat{n}}$

using the local unit normal vector $\mathbf{\hat{n}}$ (pointing towards the center of the local circle of curvature) and the local curvature of the trajectory $k$, or the local curvature radius $r = \dfrac{1}{k}$ (the radius of the local circle of curvature) so that we can recognize the more familiar centripetal acceleration in the normal acceleration.

$\dfrac{d \mathbf{v}}{dt} = \mathbf{a}_t + \mathbf{a}_n = \dfrac{ dv}{dt} \mathbf{\hat{t}} + v \dfrac{d \mathbf{\hat{t}}}{dt} = \underbrace{\dfrac{ dv}{dt}}_{a_t} \mathbf{\hat{t}} + \underbrace{k v^2}_{a_n} \mathbf{\hat{n}}$.

Short answer. Tangential acceleration makes the magnitude of the velocity change, while normal acceleration makes the direction of the velocity change.

Some details. The velocity of a point is tangent to its trajectory, and thus we can write the velocity vector as

$\mathbf{v} = v \mathbf{\hat{t}}$,

i.e. as the product of a unit vector $\mathbf{\hat{t}}$, that provides the information about the direction of the vector, and a scalar $v$, the magnitude of the velocity (the number you read on the tachometer of your car, as an example).

Now, we can take time derivative of the velocity, using the law for the derivative of a product (both the magnitude and the direction of the velocity depends on time, in general),

$\dfrac{d \mathbf{v}}{dt} = \dfrac{d}{dt} (v \mathbf{\hat{t}}) = \underbrace{\dfrac{ dv}{dt} \mathbf{\hat{t}}}_{\mathbf{a}_t} + \underbrace{v \dfrac{d \mathbf{\hat{t}}}{dt}}_{\mathbf{a}_n}$,

being the first term tangent to the trajectory (since it has the same direction as $\mathbf{\hat{t}}$), and the second term orthogonal to that, since it's in the direction of the time derivative of a unit vector, that is always orthogonal to the direction itself (or zero, if the direction doesn't change), as we can prove using

$1 = |\mathbf{\hat{t}}|^2 = \mathbf{\hat{t}} \cdot \mathbf{\hat{t}}$$\qquad \rightarrow \qquad$ $0 = \dfrac{d}{dt}|\mathbf{\hat{t}}|^2 = 2 \mathbf{\hat{t}} \cdot \dfrac{d \mathbf{\hat{t}}}{dt}$.

To get a deeper insight into normal acceleration, we need just a bit of differential geometry of curves in space. The time derivative of the unit tangent vector can be written as

$\dfrac{d \mathbf{\hat{t}}}{dt} = k v \mathbf{\hat{n}} = \frac{v}{r} \mathbf{\hat{n}}$$\qquad \rightarrow \qquad$$\mathbf{a}_n = k v^2 \mathbf{\hat{n}} = \dfrac{v^2}{r} \mathbf{\hat{n}}$

using the local unit normal vector $\mathbf{\hat{n}}$ (pointing towards the center of the local circle of curvature) and the local curvature of the trajectory $k$, or the local curvature radius $r = \frac{1}{k}$ (the radius of the local circle of curvature) so that we can recognize the more familiar centripetal acceleration in the normal acceleration. Eventually, we can write the acceleration as

$\dfrac{d \mathbf{v}}{dt} = \mathbf{a}_t + \mathbf{a}_n = \dfrac{ dv}{dt} \mathbf{\hat{t}} + v \dfrac{d \mathbf{\hat{t}}}{dt} = \underbrace{\dfrac{ dv}{dt}}_{a_t} \mathbf{\hat{t}} + \underbrace{k v^2}_{a_n} \mathbf{\hat{n}}$.

Short answer. Tangential acceleration makes the magnitude of the velocity change, while normal acceleration makes the direction of the velocity change.

Some details. The velocity of a point is tangent to its trajectory, and thus we can write the velocity vector as

$\mathbf{v} = v \mathbf{\hat{t}}$,

i.e. as the product of a unit vector $\mathbf{\hat{t}}$, that provides the information about the direction of the vector, and a scalar $v$, the magnitude of the velocity (the number you read on the tachometer of your car, as an example).

Now, we can take time derivative of the velocity, using the law for the derivative of a product (both the magnitude and the direction of the velocity depends on time, in general),

$\dfrac{d \mathbf{v}}{dt} = \dfrac{d}{dt} (v \mathbf{\hat{t}}) = \underbrace{\dfrac{ dv}{dt} \mathbf{\hat{t}}}_{\mathbf{a}_t} + \underbrace{v \dfrac{d \mathbf{\hat{t}}}{dt}}_{\mathbf{a}_n}$,

being the first term tangent to the trajectory (since it has the same direction as $\mathbf{\hat{t}}$), and the second term orthogonal to that, since it's in the direction of the time derivative of a unit vector, that is always orthogonal to the direction itself (or zero, if the direction doesn't change), as we can prove using

$1 = |\mathbf{\hat{t}}|^2 = \mathbf{\hat{t}} \cdot \mathbf{\hat{t}}$$\qquad \rightarrow \qquad$ $0 = \dfrac{d}{dt}|\mathbf{\hat{t}}|^2 = 2 \mathbf{\hat{t}} \cdot \dfrac{d \mathbf{\hat{t}}}{dt}$.

To get a deeper insight into normal acceleration, we need just a bit of differential geometry of curves in space. The time derivative of the unit tangent vector can be written as

$\dfrac{d \mathbf{\hat{t}}}{dt} = k v \mathbf{\hat{n}}$,

using the unit normal vector $\mathbf{\hat{n}}$ and the curvature of the trajectory $k$, so that we can write the acceleration as

$\dfrac{d \mathbf{v}}{dt} = \mathbf{a}_t + \mathbf{a}_n = \dfrac{ dv}{dt} \mathbf{\hat{t}} + v \dfrac{d \mathbf{\hat{t}}}{dt} = \underbrace{\dfrac{ dv}{dt}}_{a_t} \mathbf{\hat{t}} + \underbrace{k v^2}_{a_n} \mathbf{\hat{n}}$.

Short answer. Tangential acceleration makes the magnitude of the velocity change, while normal acceleration makes the direction of the velocity change.

Some details. The velocity of a point is tangent to its trajectory, and thus we can write the velocity vector as

$\mathbf{v} = v \mathbf{\hat{t}}$,

i.e. as the product of a unit vector $\mathbf{\hat{t}}$, that provides the information about the direction of the vector, and a scalar $v$, the magnitude of the velocity (the number you read on the tachometer of your car, as an example).

Now, we can take time derivative of the velocity, using the law for the derivative of a product (both the magnitude and the direction of the velocity depends on time, in general),

$\dfrac{d \mathbf{v}}{dt} = \dfrac{d}{dt} (v \mathbf{\hat{t}}) = \underbrace{\dfrac{ dv}{dt} \mathbf{\hat{t}}}_{\mathbf{a}_t} + \underbrace{v \dfrac{d \mathbf{\hat{t}}}{dt}}_{\mathbf{a}_n}$,

being the first term tangent to the trajectory (since it has the same direction as $\mathbf{\hat{t}}$), and the second term orthogonal to that, since it's in the direction of the time derivative of a unit vector, that is always orthogonal to the direction itself (or zero, if the direction doesn't change), as we can prove using

$1 = |\mathbf{\hat{t}}|^2 = \mathbf{\hat{t}} \cdot \mathbf{\hat{t}}$$\qquad \rightarrow \qquad$ $0 = \dfrac{d}{dt}|\mathbf{\hat{t}}|^2 = 2 \mathbf{\hat{t}} \cdot \dfrac{d \mathbf{\hat{t}}}{dt}$.

To get a deeper insight into normal acceleration, we need just a bit of differential geometry of curves in space. The time derivative of the unit tangent vector can be written as

$\dfrac{d \mathbf{\hat{t}}}{dt} = k v \mathbf{\hat{n}} = \frac{v}{r} \mathbf{\hat{n}}$$\qquad \rightarrow \qquad$$\mathbf{a}_n = k v^2 \mathbf{\hat{n}} = \dfrac{v^2}{r} \mathbf{\hat{n}}$

using the local unit normal vector $\mathbf{\hat{n}}$ (pointing towards the center of the local circle of curvature) and the local curvature of the trajectory $k$, or the local curvature radius $r = \dfrac{1}{k}$ (the radius of the local circle of curvature) so that we can recognize the more familiar centripetal acceleration in the normal acceleration.

$\dfrac{d \mathbf{v}}{dt} = \mathbf{a}_t + \mathbf{a}_n = \dfrac{ dv}{dt} \mathbf{\hat{t}} + v \dfrac{d \mathbf{\hat{t}}}{dt} = \underbrace{\dfrac{ dv}{dt}}_{a_t} \mathbf{\hat{t}} + \underbrace{k v^2}_{a_n} \mathbf{\hat{n}}$.