Motion question (1 Viewer)

BlueGas · Jul 1, 2015

I need help with question 10. b), thanks.

InteGrand · Jul 1, 2015

BlueGas said:
I need help with question 10. b), thanks.

$10. (b) $x(t)=12t-t^3\Rightarrow v(t) = 12-3t^2$ (differentiating).$

$i.e. $v(t)=3(4-t^2)$$

$So we can see that for $t\in [0,2)$, $v> 0$, and for $t\geq 2$, $v\leq 0$ (equality only at $t=2$). This means that the particle is moving away from $O$ for the first 2 seconds, stops instantaneously at $t=2$, and thereafter is moving towards the origin.$

$The particle returns to $O$ when $x=0$ and $t>0$, i.e. at the time such that $12t-t^3=0,t> 0 \Rightarrow 12 -t^2 = 0 \Rightarrow t^2 = 12$ (so $t=\sqrt{12}$). At this time, the velocity is $v=3(4-12)=3\times (-8)=-24$, as required.$

Drsoccerball · Jul 1, 2015

Damn it you beat me to it

Motion question (1 Viewer)

BlueGas

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InteGrand

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Drsoccerball

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