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Question Number 14668 by tawa tawa last updated on 03/Jun/17

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

$$\mathrm{please}\mathrm{the}\mathrm{two}\mathrm{questions}.$$

Commented by mrW1 last updated on 03/Jun/17

$$pleasecheckthequestion!$$$$-15t+40isnorelation!ifitmeans$$$$x=-15t+40$$$$then{it}^{\prime }samotionwithconstant$$$$velocityv=-15m/s.$$

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

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by Tinkutara last updated on 03/Jun/17

$$x_2-x_1=15(t_2-t_1)$$$$\therefore x_6-x_4=15(6-4)=30\mathrm{m}$$$$\mathrm{Answer}\left(\mathrm{d}\right):30\mathrm{m}$$$$\mathrm{Acceleration}\mathrm{is}0.$$

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

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by mrW1 last updated on 03/Jun/17

$$Q2:$$$$x=12t^3-18t^2+2t+5$$$$v=\frac{dx}{dt}=36t^2-36t+2$$$$a=\frac{dv}{dt}=72t-36$$$$a=0=72t-36$$$$\Rightarrow t=\frac{36}{72}=\frac{1}{2}s$$$$att=\frac{1}{2}theaccelerationiszero$$$$x=12{\left(\frac{1}{2}\right)}^3-18{\left(\frac{1}{2}\right)}^2+2\left(\frac{1}{2}\right)+5=3m$$$$v=36{\left(\frac{1}{2}\right)}^2-36\left(\frac{1}{2}\right)+2=-7m/s$$

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

$$\mathrm{Am}\mathrm{really}\mathrm{grateful}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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