JEE MainPhysicsEasyMotion in 1D

The velocity time relation of an electron starting from rest is
given by v=K t, where K=2 m/s^2. The distance traversed in 3 sec is 3n
then the find the value of n.

Show answer and solution

a = \frac{dv}{dt} = k = 2m/s^{2}

S = 0 + \frac{1}{2}(2)(3)^{2} = 9m

Source: Grass-G / P-05 Physics paper 10 October.docx

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