proof that, 2as =v^2 - u^2
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Newton's first law of motion is v = u + at
→ t=( v-u) / a put this equation in Newton's second law i.e. s = ut + 1/2a(t)^2
s = u(v-u)/a + 1/2 a [(v-u)/a]^2
s =uv/a - u^2/a + 1/2[v^2 + u^2 -2uv] / a
s =[ 2uv - 2u^2 + v^2 + u^2 - 2uv] / 2a
2as = v^2 - u^2 hence proved
→ t=( v-u) / a put this equation in Newton's second law i.e. s = ut + 1/2a(t)^2
s = u(v-u)/a + 1/2 a [(v-u)/a]^2
s =uv/a - u^2/a + 1/2[v^2 + u^2 -2uv] / a
s =[ 2uv - 2u^2 + v^2 + u^2 - 2uv] / 2a
2as = v^2 - u^2 hence proved
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