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-retar

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

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

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