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Question Number 134530 by I want to learn more last updated on 04/Mar/21

The speed of a train is reduced from 80km/hr to 40km/hr in a distance of 500m on applying the brakes. (i) How much further will the train travels before coming to rest. (ii) Assuming the retardation remains constant, how long will it take to bring the train to rest after the application of the brakes?

$$\mathrm{The}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{a}\:\mathrm{train}\:\mathrm{is}\:\mathrm{reduced}\:\mathrm{from}\:\:\mathrm{80km}/\mathrm{hr}\:\:\mathrm{to}\:\:\mathrm{40km}/\mathrm{hr}\:\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{distance}\:\mathrm{of}\:\:\mathrm{500m}\:\:\mathrm{on}\:\mathrm{applying}\:\mathrm{the}\:\mathrm{brakes}. \\ $$$$\left(\mathrm{i}\right)\:\:\mathrm{How}\:\mathrm{much}\:\mathrm{further}\:\mathrm{will}\:\mathrm{the}\:\mathrm{train}\:\mathrm{travels}\:\mathrm{before}\:\mathrm{coming}\:\mathrm{to}\:\mathrm{rest}. \\ $$$$\left(\mathrm{ii}\right)\:\:\mathrm{Assuming}\:\mathrm{the}\:\mathrm{retardation}\:\mathrm{remains}\:\mathrm{constant},\:\mathrm{how}\:\mathrm{long}\:\mathrm{will}\:\mathrm{it} \\ $$$$\mathrm{take}\:\mathrm{to}\:\mathrm{bring}\:\mathrm{the}\:\mathrm{train}\:\mathrm{to}\:\mathrm{rest}\:\mathrm{after}\:\mathrm{the}\:\mathrm{application}\:\mathrm{of}\:\mathrm{the}\:\mathrm{brakes}? \\ $$

Answered by mr W last updated on 04/Mar/21

80 km/h=((80000)/(60×60))=((200)/9) m/s 40 km/h=((40000)/(60×60))=((100)/9) m/s a=((v_2 ^2 −v_1 ^2 )/(2s))=(((((100)/9))^2 −(((200)/9))^2 )/(2×500))=−((10)/(27)) m/s^2 (i): s=((v_2 ^2 −v_1 ^2 )/(2a))=((0−(((100)/9))^2 )/(2×(−((10)/(27)))))=((500)/3)=166.7 m (ii): t=((v_2 −v_1 )/a)=((0−((200)/9))/(−((10)/(27))))=60 s

$$\mathrm{80}\:{km}/{h}=\frac{\mathrm{80000}}{\mathrm{60}×\mathrm{60}}=\frac{\mathrm{200}}{\mathrm{9}}\:{m}/{s} \\ $$$$\mathrm{40}\:{km}/{h}=\frac{\mathrm{40000}}{\mathrm{60}×\mathrm{60}}=\frac{\mathrm{100}}{\mathrm{9}}\:{m}/{s} \\ $$$${a}=\frac{{v}_{\mathrm{2}} ^{\mathrm{2}} −{v}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}{s}}=\frac{\left(\frac{\mathrm{100}}{\mathrm{9}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{200}}{\mathrm{9}}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{500}}=−\frac{\mathrm{10}}{\mathrm{27}}\:{m}/{s}^{\mathrm{2}} \\ $$$$\left({i}\right): \\ $$$${s}=\frac{{v}_{\mathrm{2}} ^{\mathrm{2}} −{v}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}{a}}=\frac{\mathrm{0}−\left(\frac{\mathrm{100}}{\mathrm{9}}\right)^{\mathrm{2}} }{\mathrm{2}×\left(−\frac{\mathrm{10}}{\mathrm{27}}\right)}=\frac{\mathrm{500}}{\mathrm{3}}=\mathrm{166}.\mathrm{7}\:{m} \\ $$$$\left({ii}\right): \\ $$$${t}=\frac{{v}_{\mathrm{2}} −{v}_{\mathrm{1}} }{{a}}=\frac{\mathrm{0}−\frac{\mathrm{200}}{\mathrm{9}}}{−\frac{\mathrm{10}}{\mathrm{27}}}=\mathrm{60}\:{s} \\ $$

Commented by I want to learn more last updated on 04/Mar/21

Thanks sir, i really appreciate.

$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

Commented by I want to learn more last updated on 07/Mar/21

Sir please help me with Q134704

$$\mathrm{Sir}\:\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\:\mathrm{Q134704} \\ $$

Commented by otchereabdullai@gmail.com last updated on 12/Mar/21

fantastic

$$\mathrm{fantastic} \\ $$

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