Question #6e96b

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Question #6e96b

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Answer 1 · 2015-08-01T04:04:37

$v = 8 \sqrt{5} m/s$ $v = 8sqrt(5)"m/s"$

Explanation:

Just after bus started
For bus: Displacement, $x = (\frac{1}{2})$ $x = (1/2)$ x $(0.2)$ $(0.2)$ $t^{2}$ $t^2$
For man: Displacement, $80 + x = v t$ $80 + x = vt$

To get close to the bus:

$0.1 \cdot t^{2} = v t - 80$ $0.1 * t^2 = vt - 80$ ; eliminating $x$ $x$ from above two equations

=> $t^{2} - 10 v t + 800 = 0$ $t^2 -10vt +800 = 0$

To get the real solution of the above quadratic equation;

$10 v^{2} - 4 \cdot 1 \cdot 80 \geq 0$ $10v^2 - 4 * 1 * 80 >= 0$

This means that

$v^{2} \geq 320$ $v^2 >= 320$

so, the minimum speed will be

$\sqrt{v^{2}} = \sqrt{320} \Rightarrow v = 8 \sqrt{5} m/s$ $sqrt(v^2) = sqrt(320) => v = 8sqrt(5)"m/s"$

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Question #6e96b

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