Menu Close

A-man-runs-at-constant-speed-of-v-7m-s-If-he-is-50-7m-from-the-end-of-a-train-which-has-a-steady-acceleration-of-0-25m-s-2-Will-the-man-meet-the-train-if-no-what-is-the-min-distance-between-them-

Tinku Tara 0 Comments
Pin




Question Number 121842 by 676597498 last updated on 12/Nov/20
A man runs at constant speed of v=7m/s. If he is 50.7m from the end of a train which has a steady acceleration of 0.25m/s^2 . Will the man meet the train? if no, what is the min distance between them
$$\mathrm{A}\:\mathrm{man}\:\mathrm{runs}\:\mathrm{at}\:\mathrm{constant}\:\mathrm{speed}\:\mathrm{of} \\ $$$$\mathrm{v}=\mathrm{7m}/\mathrm{s}.\:\mathrm{If}\:\mathrm{he}\:\mathrm{is}\:\mathrm{50}.\mathrm{7m}\:\mathrm{from}\:\mathrm{the}\:\mathrm{end}\:\mathrm{of}\:\mathrm{a}\:\mathrm{train}\:\mathrm{which}\:\mathrm{has}\:\mathrm{a}\:\mathrm{steady}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{0}.\mathrm{25m}/\mathrm{s}^{\mathrm{2}} . \\ $$$$\mathrm{Will}\:\mathrm{the}\:\mathrm{man}\:\mathrm{meet}\:\mathrm{the}\:\mathrm{train}? \\ $$$$\mathrm{if}\:\mathrm{no},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{min}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{them} \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 12/Nov/20
(1/2)(0.25)t^2 +50.7=7t t^2 −56t+405.6=0 ⇒t=((56±(√(56^2 −4×405.6)))/2)
$$\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{0}.\mathrm{25}\right){t}^{\mathrm{2}} +\mathrm{50}.\mathrm{7}=\mathrm{7}{t} \\ $$$${t}^{\mathrm{2}} −\mathrm{56}{t}+\mathrm{405}.\mathrm{6}=\mathrm{0}\:\:\Rightarrow{t}=\frac{\mathrm{56}\pm\sqrt{\mathrm{56}^{\mathrm{2}} −\mathrm{4}×\mathrm{405}.\mathrm{6}}}{\mathrm{2}} \\ $$
Commented by 676597498 last updated on 12/Nov/20
i dont get it sir
$$\mathrm{i}\:\mathrm{dont}\:\mathrm{get}\:\mathrm{it}\:\mathrm{sir} \\ $$
Commented by Dwaipayan Shikari last updated on 12/Nov/20
they will meet at time ′t′ in t time train will cross S=(1/2)(0.25)t^2 metre And i will be at S_0 =7t metre (1/2)(0.25)t^2 +initial distance=7t (In order to reach)
$${they}\:{will}\:{meet}\:{at}\:{time}\:'{t}' \\ $$$${in}\:{t}\:{time}\:{train}\:{will}\:{cross}\:{S}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{0}.\mathrm{25}\right){t}^{\mathrm{2}} \:{metre} \\ $$$${And}\:{i}\:{will}\:{be}\:{at}\:{S}_{\mathrm{0}} =\mathrm{7}{t}\:{metre} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{0}.\mathrm{25}\right){t}^{\mathrm{2}} +{initial}\:{distance}=\mathrm{7}{t}\:\:\left({In}\:{order}\:{to}\:{reach}\right) \\ $$
Commented by 676597498 last updated on 12/Nov/20
thankz i now get if they didnt meet how can the min distance be found pls
$$\mathrm{thankz}\:\mathrm{i}\:\mathrm{now}\:\mathrm{get} \\ $$$$\mathrm{if}\:\mathrm{they}\:\mathrm{didnt}\:\mathrm{meet}\:\mathrm{how}\:\mathrm{can}\:\mathrm{the}\:\mathrm{min} \\ $$$$\mathrm{distance}\:\mathrm{be}\:\mathrm{found}\:\mathrm{pls} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *