Question

If I have
(70 m/s^2)/0,50m do I get the units in m/s^-2 ?

Answer 1

No, you will get \[s ^{-2}\]

Answer 2

Can that be right when it's an acceleration?

Answer 3

\[a=\frac{v-u}{t}\]
a units is \[ms ^{-2}\]

Answer 4

I use a = v^2/(2*s) --> (70 m/s^2)/(2*0,50m)

Answer 5

the 2 in the denominator don't multiply by the distance, which is 0.50m

Answer 6

\[a=\frac{70ms ^{-1}}{2s}\]will do
note that the "m" and "s" here is units

Answer 7

Thank you :)

Answer 8

welcome