Question

the displacement of a body in linear motion is given by x=kt^2, comment on its acceleration?

Answer 1

Hello user !!

Here's the answer you are looking for

The differentiation of distance (x) with respect to time is velocity.

So, for x = kt²,

$v = \frac{dx}{dt} = \frac{d(k {t}^{2}) }{dt} \\ \\ v = 2kt$


And the differentiation of velocity(v) with respect to time is acceleration.

So, for v = 2kt,

$a = \frac{dv}{dt} = \frac{d(2kt)}{dt} \\ \\ a = 2k$
Since, k is a constant, acceleration is also constant. Thus, the acceleration is uniform.


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