Question

Derive for position – time equation using calculus method.
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Answer 1

The derivation is as follows:

Explanation:

We know that rate of change of position is velocity

Therefore, at any instant if velocity is v and the positional displacement is s

Then

$v=\frac{ds}{dt}$

But we know that if initial velocity is u and acceleration is a then in time t

$v=u+at$

$\implies \frac{ds}{dt}=u+at$

$\implies ds=udt+atdt$

$\implies \int_0^s ds=u\int_0^t dt+a\int_0^t dt$

$\implies s\Bigr|_0^s=ut\Bigr|_0^t+a\frac{t^2}{2}\Bigr|_0^t$

$\implies \boxed{s=ut+\frac{1}{2}at^2}$

Hope this answer is helpful.

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