1
$\begingroup$

Can I subtract, say, for example, 600 m/s - 20 m/s/s? I am confused about this topic and there is no good information on the web.

$\endgroup$
0
5
$\begingroup$

To demonstrate why you can't do this, let's step up one derivative, and explore an anecdote about why it doesn't much sense to add distance and velocity directly.


I ask you, "How far away is your house from Jill's house?"

You answer, "It is is five kilometers plus seventeen meters per second away." ($5 \,\text{km} + 17\,\text{m/s}$.)

I was asking you for a distance. But you gave me something that was neither a distance, nor a velocity, but some garbled summation of units.

Confused by your answer, I decide to change the subject. "How fast do you think that car is driving over there?"

You answer, "It is driving ten miles per hour, minus fourteen feet." ($10 \,\text{mph} - 14\,\text{ft}$.)

Wait, what are you talking about? I wonder.

I was asking you for a speed. But you gave me something that was neither a speed, nor a distance, but some garbled summation of units.

If I ask for distance, give me a distance. If I ask for a velocity, give me a velocity.

The same principle applies for all units, and is precisely why checking your units at the end of a derivation is a great sanity check that at least you are attempting to express the type of quantity you wanted to.

$\endgroup$
2
$\begingroup$

No, you can't.m/s tells us how fast something is moving/ rate of change of distance and m/s/s tells us the rate of change of speed. we cant subtract them because these two are very different from each other.

$\endgroup$
2
$\begingroup$

There is no sensical way to add or subtract them together since they are terms that have completely different units and mean totally different things.

Velocity is the rate of change of displacement whilst acceleration is the rate of change of velocity or in other words the rate of change of the rate of change of displacement. If we think of something that is a rate of change as providing units of $s^{-1}$ then the rate of change of displacement will have units of $ms^{-1}$ and the rate of change of this velocity will have units of $ms^{-2}$.

$\endgroup$
0
$\begingroup$

In physics, we can add or subtract only those terms which have same dimensions (or SI units).

Velocity $(ms^{-1})$ and acceleration $(ms^{-2})$ are terms having different dimensions each so we can't add or subtract them from each other.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy