Question

prove that S=Vf^2-Vi^2/2a is dimensionally correct.

Answer 1

s is displacement so units are metres and dimension formula is L and for velocity units are m/sec and for acceleration units are m/sec^2 dimension : L T^-2
vf^2- vi^2 units are (m/sec)^2= m^2sec^-2
dimension : L^2T^-2
TOTAL DIMENSION = L^2 T^-2÷ L T^-2 = L SAME AS DISPLACEMENT S

Answer 2

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$s = \frac{ {vf }^{2} - {vi}^{2} }{2a}$
Here

S is displacement

Units are meters

Dimension formula = L

Acceleration units are m/sec^2

Dimension

$l = {t}^{2} - 2$
${l}^{2} - l {t}^{2} - 2 \div l {t}^{2} - 2$


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