Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com