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Question Number 9637 by tawakalitu last updated on 22/Dec/16

A body starts from rest and move with uniform acceleration of 6m/s^2 . what distance does it covered in the 3rd seconds.

$$\mathrm{A}\:\mathrm{body}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{rest}\:\mathrm{and}\:\mathrm{move}\:\mathrm{with}\: \\ $$$$\mathrm{uniform}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{6m}/\mathrm{s}^{\mathrm{2}} .\:\:\mathrm{what}\: \\ $$$$\mathrm{distance}\:\mathrm{does}\:\mathrm{it}\:\mathrm{covered}\:\mathrm{in}\:\mathrm{the}\:\mathrm{3rd}\:\mathrm{seconds}. \\ $$

Answered by ridwan balatif last updated on 22/Dec/16

O→a=6m/s^2 O ∣−−−−−−−−−−−−−−∣ t=0s t=3s S=Vo.t+(1/2)a.t^2 S=0.3+(1/2).6.3^2 S=27m

$$\mathrm{O}\rightarrow\mathrm{a}=\mathrm{6m}/\mathrm{s}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{O} \\ $$$$\mid−−−−−−−−−−−−−−\mid \\ $$$$\mathrm{t}=\mathrm{0s}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{t}=\mathrm{3s} \\ $$$$\mathrm{S}=\mathrm{Vo}.\mathrm{t}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{a}.\mathrm{t}^{\mathrm{2}} \\ $$$$\mathrm{S}=\mathrm{0}.\mathrm{3}+\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{6}.\mathrm{3}^{\mathrm{2}} \\ $$$$\mathrm{S}=\mathrm{27m} \\ $$

Commented by tawakalitu last updated on 22/Dec/16

Thanks so much sir. God bless you.

$$\mathrm{Thanks}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

Answered by mrW last updated on 22/Dec/16

total distance in the first 2 seconds: s_1 =(1/2)×6×2^2 =12m total distance in the first 3 seconds: s_2 =(1/2)×6×3^2 =27m the distance in the 3rd second: 27−12=15m

$$\mathrm{total}\:\mathrm{distance}\:\mathrm{in}\:\mathrm{the}\:\mathrm{first}\:\mathrm{2}\:\mathrm{seconds}: \\ $$$$\mathrm{s}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{6}×\mathrm{2}^{\mathrm{2}} =\mathrm{12m} \\ $$$$ \\ $$$$\mathrm{total}\:\mathrm{distance}\:\mathrm{in}\:\mathrm{the}\:\mathrm{first}\:\mathrm{3}\:\mathrm{seconds}: \\ $$$$\mathrm{s}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{6}×\mathrm{3}^{\mathrm{2}} =\mathrm{27m} \\ $$$$ \\ $$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{in}\:\mathrm{the}\:\mathrm{3rd}\:\mathrm{second}: \\ $$$$\mathrm{27}−\mathrm{12}=\mathrm{15m} \\ $$

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