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Two cars are travelling on a straight road. The car c1 is going at a constant speed of 9m/s. 40 metres ahead of c1, car c2 initially at rest, but starts moving at an acceleration of 1m/s^2.During their motion, the car c1 overtakes c2, but soon c2 overtakes c1. Determine the maximum lead that c1 can have from c2:

a) 0.5m b)9.5m c)81m d)90m

Dear Student,

Please find below the solution to the asked query:

If the separation between the C_{1} and C_{2} is maximum, then the difference of the first car displacement and the displacement of the second car displacement should have its maximum value. Therefore,

${S}_{1}=9t\phantom{\rule{0ex}{0ex}}{S}_{2}=\frac{1}{2}\times 1\times {t}^{2}\phantom{\rule{0ex}{0ex}}So,thedifferenceis,\phantom{\rule{0ex}{0ex}}\u2206S={S}_{1}-{S}_{2}=9t-\frac{{t}^{2}}{2}\phantom{\rule{0ex}{0ex}}Forthistobemaximum,\phantom{\rule{0ex}{0ex}}\frac{d\left(\u2206S\right)}{dt}=0\Rightarrow \frac{d}{dt}\left\{9t-\frac{{t}^{2}}{2}\right\}=0\Rightarrow 9-t=0\Rightarrow t=9sec.\phantom{\rule{0ex}{0ex}}So,\phantom{\rule{0ex}{0ex}}\u2206S=9\times 9-\frac{{9}^{2}}{2}=40.5m\phantom{\rule{0ex}{0ex}}Therefore,initiallythefirstcaris40mbehindthesecondcar.\phantom{\rule{0ex}{0ex}}Thereforethemaximumleadis0.5m$

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