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Question Number 121842 by 676597498 last updated on 12/Nov/20

$$\mathrm{A}\mathrm{man}\mathrm{runs}\mathrm{at}\mathrm{constant}\mathrm{speed}\mathrm{of}$$$$\mathrm{v}=7\mathrm{m}/\mathrm{s}.\mathrm{If}\mathrm{he}\mathrm{is}50.7\mathrm{m}\mathrm{from}\mathrm{the}\mathrm{end}\mathrm{of}\mathrm{a}\mathrm{train}\mathrm{which}\mathrm{has}\mathrm{a}\mathrm{steady}\mathrm{acceleration}\mathrm{of}0.25\mathrm{m}/{\mathrm{s}}^2.$$$$\mathrm{Will}\mathrm{the}\mathrm{man}\mathrm{meet}\mathrm{the}\mathrm{train}?$$$$\mathrm{if}\mathrm{no},\mathrm{what}\mathrm{is}\mathrm{the}\min \mathrm{distance}\mathrm{between}\mathrm{them}$$$$$$

Answered by Dwaipayan Shikari last updated on 12/Nov/20

$$\frac{1}{2}(0.25)t^2+50.7=7t$$$$t^2-56t+405.6=0\Rightarrow t=\frac{56\pm \sqrt{{56}^2-4\times 405.6}}{2}$$

Commented by 676597498 last updated on 12/Nov/20

$$\mathrm{i}\mathrm{dont}\mathrm{get}\mathrm{it}\mathrm{sir}$$

Commented by Dwaipayan Shikari last updated on 12/Nov/20

$$theywillmeetattime{}^{\prime }{t}^{\prime }$$$$inttimetrainwillcrossS=\frac{1}{2}(0.25)t^{2}metre$$$$Andiwillbeat{S}_0=7tmetre$$$$\frac{1}{2}(0.25)t^2+initialdistance=7t\left(Inordertoreach\right)$$

Commented by 676597498 last updated on 12/Nov/20

$$\mathrm{thankz}\mathrm{i}\mathrm{now}\mathrm{get}$$$$\mathrm{if}\mathrm{they}\mathrm{didnt}\mathrm{meet}\mathrm{how}\mathrm{can}\mathrm{the}\min$$$$\mathrm{distance}\mathrm{be}\mathrm{found}\mathrm{pls}$$

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