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Question Number 19828 by NECC last updated on 16/Aug/17

The speed of a train is reduced  from 80km/h to 40km/h after  the application of the brake.  (i)how much further would the   train travel before coming to rest  (ii)assuming the acceleration is  kept constant,how long will it  take to bring the train to rest  after the application of the brakes?

$${The}\:{speed}\:{of}\:{a}\:{train}\:{is}\:{reduced} \\ $$$${from}\:\mathrm{80}{km}/{h}\:{to}\:\mathrm{40}{km}/{h}\:{after} \\ $$$${the}\:{application}\:{of}\:{the}\:{brake}. \\ $$$$\left({i}\right){how}\:{much}\:{further}\:{would}\:{the}\: \\ $$$${train}\:{travel}\:{before}\:{coming}\:{to}\:{rest} \\ $$$$\left({ii}\right){assuming}\:{the}\:{acceleration}\:{is} \\ $$$${kept}\:{constant},{how}\:{long}\:{will}\:{it} \\ $$$${take}\:{to}\:{bring}\:{the}\:{train}\:{to}\:{rest} \\ $$$${after}\:{the}\:{application}\:{of}\:{the}\:{brakes}? \\ $$

Commented by NECC last updated on 16/Aug/17

so sorry.... The speed of a car was  reduced from 80km/h to 40km/h  for 500m.......    the other texts are given

$${so}\:{sorry}....\:{The}\:{speed}\:{of}\:{a}\:{car}\:{was} \\ $$$${reduced}\:{from}\:\mathrm{80}{km}/{h}\:{to}\:\mathrm{40}{km}/{h} \\ $$$${for}\:\mathrm{500}{m}....... \\ $$$$ \\ $$$${the}\:{other}\:{texts}\:{are}\:{given} \\ $$$$ \\ $$

Answered by Tinkutara last updated on 16/Aug/17

v^2  − u^2  = 2as  a = ((40^2  − 80^2 )/(2×0.5)) = −4800 km/h^2   (i) s = ((v^2  − u^2 )/(2a)) = ((0 − 1600)/(−2×4800)) = (1/6) km  Means, (1/6) km more after travelling  500 m already.  (ii) t = ((v − u)/a) = ((0 − 80)/(−4800)) = (1/(60)) h = 1 min

$${v}^{\mathrm{2}} \:−\:{u}^{\mathrm{2}} \:=\:\mathrm{2}{as} \\ $$$${a}\:=\:\frac{\mathrm{40}^{\mathrm{2}} \:−\:\mathrm{80}^{\mathrm{2}} }{\mathrm{2}×\mathrm{0}.\mathrm{5}}\:=\:−\mathrm{4800}\:\mathrm{km}/\mathrm{h}^{\mathrm{2}} \\ $$$$\left({i}\right)\:{s}\:=\:\frac{{v}^{\mathrm{2}} \:−\:{u}^{\mathrm{2}} }{\mathrm{2}{a}}\:=\:\frac{\mathrm{0}\:−\:\mathrm{1600}}{−\mathrm{2}×\mathrm{4800}}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\:\mathrm{km} \\ $$$$\mathrm{Means},\:\frac{\mathrm{1}}{\mathrm{6}}\:\mathrm{km}\:\mathrm{more}\:\mathrm{after}\:\mathrm{travelling} \\ $$$$\mathrm{500}\:\mathrm{m}\:\mathrm{already}. \\ $$$$\left({ii}\right)\:{t}\:=\:\frac{{v}\:−\:{u}}{{a}}\:=\:\frac{\mathrm{0}\:−\:\mathrm{80}}{−\mathrm{4800}}\:=\:\frac{\mathrm{1}}{\mathrm{60}}\:\mathrm{h}\:=\:\mathrm{1}\:\mathrm{min} \\ $$

Commented by NECC last updated on 16/Aug/17

thanks boss...  please do you know any site or   pdf file that really explains the  equation of motion

$${thanks}\:{boss}... \\ $$$${please}\:{do}\:{you}\:{know}\:{any}\:{site}\:{or}\: \\ $$$${pdf}\:{file}\:{that}\:{really}\:{explains}\:{the} \\ $$$${equation}\:{of}\:{motion} \\ $$$$ \\ $$$$ \\ $$

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