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Question Number 191091 by otchereabdullai last updated on 18/Apr/23
The front of a train 80m long passes  a signal at a speed of 72km/h. If the  rear of the train passes the signal   5seconds later, find   (a) the acceleration of the train  (b) the speed at which the rear of the  train passes the signal.
$${The}\:{front}\:{of}\:{a}\:{train}\:\mathrm{80}{m}\:{long}\:{passes} \\ $$$${a}\:{signal}\:{at}\:{a}\:{speed}\:{of}\:\mathrm{72}{km}/{h}.\:{If}\:{the} \\ $$$${rear}\:{of}\:{the}\:{train}\:{passes}\:{the}\:{signal}\: \\ $$$$\mathrm{5}{seconds}\:{later},\:{find}\: \\ $$$$\left({a}\right)\:{the}\:{acceleration}\:{of}\:{the}\:{train} \\ $$$$\left({b}\right)\:{the}\:{speed}\:{at}\:{which}\:{the}\:{rear}\:{of}\:{the} \\ $$$${train}\:{passes}\:{the}\:{signal}. \\ $$
Answered by mr W last updated on 18/Apr/23
v_1 =72 km/h=20 m/s  s=v_1 t+((at^2 )/2)  80=20×5+((a×5^2 )/2)  ⇒a=−(8/5)=−1.6 m/s^2   v_2 =v_1 +at=20−1.6×5=12 m/s=43.2 km/s
$${v}_{\mathrm{1}} =\mathrm{72}\:{km}/{h}=\mathrm{20}\:{m}/{s} \\ $$$${s}={v}_{\mathrm{1}} {t}+\frac{{at}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{80}=\mathrm{20}×\mathrm{5}+\frac{{a}×\mathrm{5}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{a}=−\frac{\mathrm{8}}{\mathrm{5}}=−\mathrm{1}.\mathrm{6}\:{m}/{s}^{\mathrm{2}} \\ $$$${v}_{\mathrm{2}} ={v}_{\mathrm{1}} +{at}=\mathrm{20}−\mathrm{1}.\mathrm{6}×\mathrm{5}=\mathrm{12}\:{m}/{s}=\mathrm{43}.\mathrm{2}\:{km}/{s} \\ $$
Commented by otchereabdullai last updated on 18/Apr/23
May Allah almighty bless you
$${May}\:{Allah}\:{almighty}\:{bless}\:{you} \\ $$

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