Why it is not always true that $\frac{d}{dt} |v| = |a|$ where $v$ and $a$ are respectively velocity and acceleration vectors

Question

As stated in the title. I can't understand how this can be true. For example if we take $v = 1i + 2tj$, then $a = 2j$. Size of the $a$ is $2$ while $\frac{d}{dt}|v| = \frac{d}{dt}\sqrt{4t^2 + 1} \ne 2$. Why the size of acceleration vector not equal to the derivative of speed ($|v|$)?

Answer 1

Since$$\frac{d}{dt}|v|=\frac{1}{2|v|}\frac{d}{dt}\sum_iv_i^2=\frac{\sum_iv_ia_i}{|v|}=\frac{v\cdot a}{|v|},$$a trajectory satisfies $\frac{d}{dt}|v|=|a|$ iff $v\cdot a=|v||a|$, i.e. iff $v$ is parallel to $a$.

Many situations deviate from this. For example, when acceleration is due to a radial force but the initial velocity is neither zero nor radial, the velocity won't be parallel to acceleration. This is why gravity, which we think of in everyday life as making things fall, can also cause orbits: planets don't fall into the Sun because their velocity isn't parallel to their acceleration (in fact, the two are almost perpendicular).

Answer 2

There are two ways the velocity can change.

1. You can change your speed.

2. You can change the direction you are moving in.

The first type of change is equal to $\frac{d|v|}{dt}$ and this corresponds to the component of your acceleration vector which is parallel to the direction of motion.

The second type of change is due to the component of your acceleration vector which is perpendicular to the direction of motion. It is possible to change the direction you are moving in without change your speed (e.g. uniform circular motion).

Answer 3

Consider a planet moving in a perfect circle about a star. The speed $|v|$ is constant, so $\frac{d}{dt}|v|=0$. And the absolute value $|a|$ of the acceleration is also constant, but non-zero. So $\frac{d}{dt}|v|\ne|a|$.