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




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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