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

Answered by Tinkutara last updated on 03/Jun/17

$$x=6t^3-8t^2-15t+40$$$$v=18t^2-16t-15$$$$\left(a\right)18t^2-16t-15=0$$$$t=\frac{16\pm 2\sqrt{334}}{36}=\frac{8\pm \sqrt{334}}{18}\approx 1.46\mathrm{s}$$$$\left(b\right)x(1.46)=19.72$$$$x\left(0\right)=40$$$$\mathrm{Position}\mathrm{at}t=1.46\mathrm{s}\mathrm{is}19.72$$$$\mathrm{Distance}\mathrm{travelled}\mathrm{in}\mathrm{that}\mathrm{time}=20.28\mathrm{m}$$$$\left(c\right)a=36t-16$$$$a(1.46)=36.56\mathrm{m}/{\mathrm{s}}^2$$$$\left(d\right)x\left(6\right)=958,x\left(4\right)=236$$$$\mathrm{Answer}:722\mathrm{m}$$

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

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

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