Why is acceleration inversely proportional to mass?

an object moving at a velocity of x carries the force of its mass times its speed.

when you apply a force onto an object, the increase in speed of it would be affected by its mass. Think of it this way: you apply some force onto an iron ball, and apply the same force on a plastic ball (they are of equal volume). Which one moves faster, and which one moves slower? The answer is obvious: the iron ball will accelerate slower and travel slower, while the plastic ball is faster.

The iron ball has a greater mass, so the force which makes it accelerate is deduced more. The plastic ball has a smaller mass, so the force applied is divided by a smaller number.

I hope this helps you a bit.

Say we wish to keep a force `F` exerted by an object constant. If the mass `m` of the object doubles, what must happen to the object's acceleration `a` to keep `F` unchanged?

The answer is: the object's acceleration must be halved.

We start with

`F=m*a`

and if we double the mass to `2m`, the RHS as a whole has doubled. Thus, the LHS also doubles, meaning we get double the force:

`2F = 2m*a`

This is an example of direct proportionality between `F` and `m`. If `m` doubles, `F` responds by doubling as well.

But we want to keep the force the same; we don't want `2F`, we want `F`. So we need to divide the LHS by 2. And to do that, we must divide the RHS by 2 as well. So either the mass `2m` goes back down to `m`, or the acceleration `a` gets cut to `1/2 a`.

`F= 2m*1/2 a`

This is an example of inverse proportionality. When the force is taken as a constant, if mass doubles, acceleration must be halved.

Note:

You can also see the inverse relation between `m` and `a` by solving `F=ma` for one or the other.

`F=ma " "=>" "a=F/m" " <=>" "a=F(m^-1)`

`color(white)(F=ma) " "=>" "m=F/a" "<=>" "m=F(a^-1)`

It's now easy to see mathematically that `a` and `m` are inversely proportional, because each is a multiple of the other's inverse (that multiple being `F` itself).