# How is acceleration measured if there is change only in the direction of velocity & not magnitude?

The formal definition of acceleration is

It is the measure of how fast the velocity changes with respect to time.

Now, when the magnitude of velocity changes, acceleration $$\vec{a} = \frac{d \vec{v}}{dt}$$ and its unit is $m.s^{-2}$ .

But what about the case where the velocity only changes its direction keeping its magnitude constant? The definition can then be

Acceleration is the measure of how fast the velocity changes its direction with respect to time.

But how can I calculate the acceleration then?? What should be the unit then??? Please help.

## 4 Answers

The change in velocity can be calculated by vector subtraction. ($d\vec{v} = \vec{v_f} - \vec{v_i}$).

Divide by the time between the two velocities to generate an acceleration. The direction of the acceleration will be the same as the direction of the difference vector. The magnitude of the acceleration will be the same as the magnitude of the difference vector divided by the time.

Example, at time $t=0$ seconds the velocity is in the $x$ direction, and at time $t=2$ seconds the velocity is in the y direction. At both times the velocity is 1 m/s: $$\vec{v(0)} = (1,0,0) ~\mathrm{m/s}$$ $$\vec{v(2)} = (0,1,0) ~\mathrm{m/s}$$ $$d\vec{v} = \vec{v(2)} - \vec{v(0)}$$ $$d\vec{v} = (-1,1,0) ~\mathrm{m/s}$$ $$dt = 2 - 0 = 2 ~\mathrm{s}$$ $$\vec{a} = \frac{d\vec{v}}{dt} = (-\frac12,\frac12,0) ~\mathrm{m/s^2}$$

But what about the case where the velocity only changes its direction keeping its magnitude constant?

The first equation you give is a vector equation which, in this case, means there are three equations:

$$\vec a = \frac{d\vec v}{dt}\Rightarrow a_x = \frac{dv_x}{dt}; a_y = \frac{dv_y}{dt}; a_z = \frac{dv_z}{dt}$$

If we have the additional constraint that the magnitude (squared) of the velocity is constant

$$\frac{d}{dt} (v^2_x + v^2_y + v^2_z)= 0 = \frac{d}{dt}\vec v \cdot \vec v$$

It follows that

$$v_x a_x + v_y a_y + v_z a_z = 0 = \vec v \cdot \vec a$$

In other words, there is no difference in the calculation or unit of acceleration in the case that the speed is constant.

Rather, there a constraint on the components of acceleration - the (time changing) acceleration vector must be orthogonal to the (time changing) velocity vector.

To emphasize, in the general case of acceleration, both the magnitude and direction of the velocity change with time.

But, even in the special case that only the direction of the velocity changes with time, the velocity is still changing with time thus, the formal definition always holds.

Since both velocity and acceleration are vectors, it is possible as in the case of circular motion for the magnitude of the velocity vector to remain constant but for the direction of the vector to change. Since acceleration is the rate of change of the velocity vector, acceleration would be non-zero even if the velocity vector magnitude is constant, but direction changes. For circular motion specifically this link shows the derivation of the equation $a = v^2/r$ by simply taking the double derivative of the position vector $$\vec{r} = r.cos \omega .t .\hat{i} + rsin\omega .t .\hat{j}$$

There’s a slight issue with some of your wording. Measuring and calculating a quantity are two very distinct tasks - one is experimental in nature, and one lies upon mathematical formalism and theory. That is not to say that they are entirely separate, but it’s worth noting from a pedagogical standpoint.

Fundamentally, the answer to your question relies on the notion that kinematics vectors, be they accelerations, velocities, positions, etc, all can be resolved into components which can then be measured or calculated independently of the other components. The way in which we resolve these vectors is with respect to a particular basis. That which we are most familiar with is the Cartesian basis, and is essentially the $x,y,z$ coordinate axes in $\mathbb{R}^3$.

So, we might have a velocity vector given in the following form,

$\vec{v} = v_x \hat{x} + v_y \hat{y} + v_z \hat{z}$,

wherein $\hat{x},\hat{y},\hat{z}$ just represent the $x,y,z$ directions. We have the speed defined by:

$v = \sqrt{v_x^2 + v_y^2 + v_z^2}$,

and you may convince yourself that we can change a particular component of $\vec{v}$ without changing the value of the speed, $v$.

So, the way that we would measure or calculate this quantities if we have unchanging speed but changing velocity is simply to measure or calculate them independently of the other vectors. It’s the same exact type of problem as your $2$ dimensional kinematics problems wherein you throw a ball off a cliff and have to calculate the initial and final speed… your analysis simply treats the $x$ and $y$ (horizontal and vertical) directions independently of one another.

I hope that’s helpful!