# Quick clarification on equation of motion

A 900N force acts on a 5kg box in horizontal direction initially at rest. When V=6m/s the force stops.
Through what distance was the block pushed.

I said this:
F=ma
v=x/t ---> t=x/v (1)
a=v/t (2)
combine (1) and (2)
a=v^2/x
So x=v^2/a

That is the same as angular circular motion.

Then there is
x=0.5at^2
With t=x/v
x=0.5a(x/v)^2
1=0.5ax/v^2
2=ax/v^2
x=(2v^2)/a

Where am I screwing up?

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v = x/t is valid only for non-accelerated motion, which is not the case here.

a = v/t is correct.

Thank you for your response Voko :)

I'm an idiot.
This is why I feel hesitant to help people on short notice (after being woken up no less!) because I get all overwhelmed and want to make sure everything is perfect and my mind starts making stupid mistakes.

So the equation x=(2v^2)/a needs to be reworked because I used v=x/t again.

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so x=v^2/(2a)..

x=(at^2)/2

so t=v/a

So for going up an incline with friction,

It has an initial known KE, thus v

a comes from Fnet=ma

Fnet=u(mg)cos(60)+(mg)cos(60)

Then x=v^2/(2a)
So x is the displacement up the incline correct?

Why is mind getting on the fact that velocity is positive and a is negative? but displacement shouldn't be negative

The formula you derived in #4 is valid for motion from rest till some given velocity. In #5, however, it is the other way around.

The formula you derived in #4 is valid for motion from rest till some given velocity. In #5, however, it is the other way around.
I'm sorry I don't know what you mean by other way around.

But for #5 up the incline, wouldn't it be x=[(vf2)-(vi2)]/(2a)

So negative v on top, negative a on bottom. That takes care of the negative.

But I still don't understand if this can be used to find x up an incline. What is the other way around?

The other way around is you start with non-zero velocity, and end up with no velocity. The new formula you have is the correct one, it gives you the correct sign of the displacement.

Ok I see what you're saying now..

Thank you very much!