Show that v2 +u2=2as is dimension correct with method

Answer 1

To show that the equation v^2 + u^2 = 2as is dimensionally correct, we need to check that both sides of the equation have dimensions that are consistent with each other.

Let's break down the dimensions of each term in the equation:

v^2:
- v has dimensions of length/time (L/T)
- Therefore, v^2 has dimensions of (L/T)^2 = L^2/T^2

u^2:
- u has dimensions of length/time (L/T)
- Therefore, u^2 has dimensions of (L/T)^2 = L^2/T^2

2as:
- a has dimensions of length/time^2 (L/T^2)
- s has dimensions of length (L)
- Therefore, 2as has dimensions of 2(L/T^2)*L = 2L^2/T^2

Now, we can compare the dimensions of both sides of the equation:

v^2 + u^2 = 2as
(L^2/T^2) + (L^2/T^2) = 2(L^2/T^2)

Both sides have dimensions of L^2/T^2, which means the equation is dimensionally correct.