September 2019 – Physicspupil

(bad example)

The distance between $X$ and $Z$ is $s=ut+\frac{1}{2}at^2$ .

The mid-point $Y$ of $XZ$ is $\frac{1}{2}s$ away from $X$ after some time $t'<t$ .

$\begin{aligned} \frac{1}{2}s & =ut'+\frac{1}{2}at'^2 \\ \frac{1}{2}\bigg( ut+\frac{1}{2}at^2 \bigg) & =ut'+\frac{1}{2}at'^2 \\ 0 & = \bigg( \frac{1}{2}a\bigg) t'^2 + (u)t' + \bigg( -\frac{1}{2}ut-\frac{1}{4}at^2 \bigg) \\ t' & = \frac{-(u)\pm \sqrt{(u)^2-4\big( \frac{1}{2}a\big)\big( -\frac{1}{2}ut-\frac{1}{4}at^2\big)}}{2\big(\frac{1}{2}a\big)} \\ t' & = \frac{-u\pm \sqrt{u^2+aut+\frac{1}{2}a^2t^2}}{a} \\ \dots\textrm{ because } &v-u=at\enspace \dots \\ t' & = \frac{-u\pm \sqrt{u^2+u(v-u)+\frac{1}{2}(v-u)^2}}{a} \\ & = \frac{-u\pm \sqrt{\displaystyle{\frac{u^2+v^2}{2}}}}{a} \end{aligned}$

At mid-point $Y$ the speed $w$ should therefore be

$\begin{aligned} w & =u+at' \\ & = u + a \Bigg( \frac{-u\pm \sqrt{\frac{u^2+v^2}{2}}}{a} \Bigg) \\ & = \pm \sqrt{\displaystyle{\frac{u^2+v^2}{2}}} \end{aligned}$

And the answer is A.

Remark.

What above was done brutally and mechanically, without a sense of physical meaning.

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Month: September 2019

201909160458 Solution to 2001-AL-PHY-IIA-5