Question-14692

Question Number 14692 by tawa tawa last updated on 03/Jun/17

Answered by Tinkutara last updated on 03/Jun/17

x = 6t^3 − 8t^2 − 15t + 40 v = 18t^2 − 16t − 15 (a) 18t^2 − 16t − 15 = 0 t = ((16 ± 2(√(334)))/(36)) = ((8 ± (√(334)))/(18)) ≈ 1.46 s (b) x(1.46) = 19.72 x(0) = 40 Position at t = 1.46 s is 19.72 Distance travelled in that time = 20.28 m (c) a = 36t − 16 a(1.46) = 36.56 m/s^2 (d) x(6) = 958, x(4) = 236 Answer: 722 m

$${x}\:=\:\mathrm{6}{t}^{\mathrm{3}} \:−\:\mathrm{8}{t}^{\mathrm{2}} \:−\:\mathrm{15}{t}\:+\:\mathrm{40} \\ $$$${v}\:=\:\mathrm{18}{t}^{\mathrm{2}} \:−\:\mathrm{16}{t}\:−\:\mathrm{15} \\ $$$$\left({a}\right)\:\mathrm{18}{t}^{\mathrm{2}} \:−\:\mathrm{16}{t}\:−\:\mathrm{15}\:=\:\mathrm{0} \\ $$$${t}\:=\:\frac{\mathrm{16}\:\pm\:\mathrm{2}\sqrt{\mathrm{334}}}{\mathrm{36}}\:=\:\frac{\mathrm{8}\:\pm\:\sqrt{\mathrm{334}}}{\mathrm{18}}\:\approx\:\mathrm{1}.\mathrm{46}\:\mathrm{s} \\ $$$$\left({b}\right)\:{x}\left(\mathrm{1}.\mathrm{46}\right)\:=\:\mathrm{19}.\mathrm{72} \\ $$$${x}\left(\mathrm{0}\right)\:=\:\mathrm{40} \\ $$$$\mathrm{Position}\:\mathrm{at}\:{t}\:=\:\mathrm{1}.\mathrm{46}\:\mathrm{s}\:\mathrm{is}\:\mathrm{19}.\mathrm{72} \\ $$$$\mathrm{Distance}\:\mathrm{travelled}\:\mathrm{in}\:\mathrm{that}\:\mathrm{time}\:=\:\mathrm{20}.\mathrm{28}\:\mathrm{m} \\ $$$$\left({c}\right)\:{a}\:=\:\mathrm{36}{t}\:−\:\mathrm{16} \\ $$$${a}\left(\mathrm{1}.\mathrm{46}\right)\:=\:\mathrm{36}.\mathrm{56}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \\ $$$$\left({d}\right)\:{x}\left(\mathrm{6}\right)\:=\:\mathrm{958},\:{x}\left(\mathrm{4}\right)\:=\:\mathrm{236} \\ $$$$\mathrm{Answer}:\:\mathrm{722}\:\mathrm{m} \\ $$

Commented by tawa tawa last updated on 03/Jun/17

God bless you sir. Thanks for your effort.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{effort}. \\ $$

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