Question

A particle is moving in a straight line has initial velocity 14m/s and retardation of 4m/s square . Find the distance and displacement in the 4th second.​

Answer 1

Answer:

$s = ut + 0.5a {t }^{2} \\ s = 14 \times 4 - 0.5 \times 16 \times 4 \\ = 56 - 32 = 24m$

Explanation:

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