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

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

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

$$80km/h=\frac{80000}{60\times 60}=\frac{200}{9}m/s$$$$40km/h=\frac{40000}{60\times 60}=\frac{100}{9}m/s$$$$a=\frac{v_2^2-v_1^2}{2s}=\frac{{\left(\frac{100}{9}\right)}^2-{\left(\frac{200}{9}\right)}^2}{2\times 500}=-\frac{10}{27}m/s^2$$$$\left(i\right):$$$$s=\frac{v_2^2-v_1^2}{2a}=\frac{0-{\left(\frac{100}{9}\right)}^2}{2\times (-\frac{10}{27})}=\frac{500}{3}=166.7m$$$$\left(ii\right):$$$$t=\frac{v_2-v_1}{a}=\frac{0-\frac{200}{9}}{-\frac{10}{27}}=60s$$

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

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

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

$$\mathrm{Sir}\mathrm{please}\mathrm{help}\mathrm{me}\mathrm{with}\mathrm{Q}134704$$

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

$$\mathrm{fantastic}$$

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