Question

test dimensionally if the equation v^2 = u^2 +2ax may be correct?

Answer 1

you need to square the dimensions used for velocity
you need to aggregate the dimensions used for acceleration and displacement

presuming you are using metric, you only need to think about metres & seconds, so it should be fairly straightforward for you to do.

Answer 2

v in dimension=L/T=u
s=L
a=L/T^2
L^2/T^2 = L^2/T^2 + 2 (L/T^2)(L)
0=2(L^2/T2)
0=0
hence equation is correct

Answer 3

bingo!!!

Answer 4

Checking the equation:
v^2 = [LT^-1]^2 = [L^2T^-2]
u^2 = [LT^-1]^2 = [L^2T^-2]
2ax = [LT^-2][L] = [L^2T^-2]
Since all the physical quantities have same dimensions,
Therefore the equation is dimensionally correct.

Answer 5

shivi can you please explain what you did about 2ax

Answer 6

yea sure!
Since 2 is a pure number, therefore it is dimensionless.
a i.e. acceleration has dimension [LT^-2]
x i.e. distance has dimension [L]
So. [2ax] = [LT^-2] * [L]
= [L^2T^-2]
hope it helps! :)

Answer 7

thanks shivi...