# Velocity divided by Acceleration gives distance?

See I figured that since Velocity = m/s
Acceleration = m/s^2

If I have velocity divided by Acceleration
----> m/s ÷ m/s^2 = s

Relevant equations

Velocity --> s/t
Acceleration --> (v-u)/t

The attempt at a solution
My idea seems reasonable to me but somehow I couldn't apply this logic to related questions. Based on my understanding, velocity divided by acceleration gives distance as 's' but it don't seems applicable when I attempted questions with this approach.

billy_joule
. Based on my understanding, velocity divided by acceleration gives distance as 's' .

No it doesn't. You're mixing symbols with units, s is short for seconds. eg velocity is metres per second (m/s)
In the kinematic equations (SUVAT) 's' is used to represent distance (which has units of metres)

s is distance in metres (m)
u is initial velocity in metres per second (m/s)
v is final velocity in metres per second (m/s)
a is acceleration in metres per second squared (m/s2)
t is time in seconds (s)

Mongster
Oh wait... I see the mistake now oh my, hahaha! It is really stupid... *cringing*
But thanks alot for the detailed explanation there, appreciate it really!

Cheers!

SteamKing
Staff Emeritus
Homework Helper
See I figured that since Velocity = m/s
Acceleration = m/s^2

If I have velocity divided by Acceleration
----> m/s ÷ m/s^2 = s

Relevant equations

Velocity --> s/t
Acceleration --> (v-u)/t

The attempt at a solution
My idea seems reasonable to me but somehow I couldn't apply this logic to related questions. Based on my understanding, velocity divided by acceleration gives distance as 's' but it don't seems applicable when I attempted questions with this approach.
That's only because the masses (m) canceled out.