Question Number 45085 by Necxx last updated on 08/Oct/18
$${A}\:{motorist}\:{travelled}\:{from}\:{A}\:{to}\:{B}. \\ $$$${This}\:{is}\:{a}\:{distance}\:{of}\:\mathrm{142}{km}\:{at}\:{an} \\ $$$${average}\:{speed}\:{of}\:\mathrm{60}{kmhr}^{−\mathrm{1}} .{He} \\ $$$${spent}\:\mathrm{5}/\mathrm{2}{hours}\:{in}\:{B}\:{and}\:{then} \\ $$$${returned}\:{to}\:{A}\:{at}\:{an}\:{average}\:{speed} \\ $$$${of}\:\mathrm{80}{kmh}^{−\mathrm{1}} . \\ $$$$\left.{a}\right){At}\:{what}\:{time}\:{did}\:{the}\:{man}\:{arrive} \\ $$$${back}\:{at}\:{A} \\ $$$$\left.{b}\right){find}\:{the}\:{average}\:{speed}\:{for}\:{the}_{} \\ $$$${total}\:{journey}. \\ $$
Answered by MJS last updated on 08/Oct/18
![journey starts at A at time=0 arrival in B after ((142km)/(60km hr^(−1) ))=((71)/(30))hr=2hr22min time=2hr22min 5hr30 min in B, start from B at 7hr52min arrival in A after ((142km)/(80km hr^(−1) ))=((71)/(40))hr=1hr46min30s time=9hr38min30s the average speed (without the 5:30 rest) is ((2×142km)/((((71)/(30))+((71)/(40)))hr))=((480)/7)km hr^(−1) ≈68.57km hr^(−1) [with the rest included it′s ((2×142km)/(9hr38min30s))= =((34080)/(1157))km hr^(−1) ≈29.46km hr^(−1) ]](https://www.tinkutara.com/question/Q45095.png)
$$\mathrm{journey}\:\mathrm{starts}\:\mathrm{at}\:{A}\:\mathrm{at}\:\mathrm{time}=\mathrm{0} \\ $$$$\mathrm{arrival}\:\mathrm{in}\:{B}\:\mathrm{after}\:\frac{\mathrm{142km}}{\mathrm{60km}\:\mathrm{hr}^{−\mathrm{1}} }=\frac{\mathrm{71}}{\mathrm{30}}\mathrm{hr}=\mathrm{2hr22min} \\ $$$$\:\:\:\:\:\mathrm{time}=\mathrm{2hr22min} \\ $$$$\mathrm{5hr30}\:\mathrm{min}\:\mathrm{in}\:{B},\:\mathrm{start}\:\mathrm{from}\:{B}\:\mathrm{at}\:\mathrm{7hr52min} \\ $$$$\mathrm{arrival}\:\mathrm{in}\:{A}\:\mathrm{after}\:\frac{\mathrm{142km}}{\mathrm{80km}\:\mathrm{hr}^{−\mathrm{1}} }=\frac{\mathrm{71}}{\mathrm{40}}\mathrm{hr}=\mathrm{1hr46min30s} \\ $$$$\:\:\:\:\:\mathrm{time}=\mathrm{9hr38min30s} \\ $$$$ \\ $$$$\mathrm{the}\:\mathrm{average}\:\mathrm{speed}\:\left(\mathrm{without}\:\mathrm{the}\:\mathrm{5}:\mathrm{30}\:\mathrm{rest}\right)\:\mathrm{is} \\ $$$$\frac{\mathrm{2}×\mathrm{142km}}{\left(\frac{\mathrm{71}}{\mathrm{30}}+\frac{\mathrm{71}}{\mathrm{40}}\right)\mathrm{hr}}=\frac{\mathrm{480}}{\mathrm{7}}\mathrm{km}\:\mathrm{hr}^{−\mathrm{1}} \approx\mathrm{68}.\mathrm{57km}\:\mathrm{hr}^{−\mathrm{1}} \\ $$$$\:\:\:\:\:\left[\mathrm{with}\:\mathrm{the}\:\mathrm{rest}\:\mathrm{included}\:\mathrm{it}'\mathrm{s}\:\frac{\mathrm{2}×\mathrm{142km}}{\mathrm{9hr38min30s}}=\right. \\ $$$$\left.\:\:\:\:\:\:=\frac{\mathrm{34080}}{\mathrm{1157}}\mathrm{km}\:\mathrm{hr}^{−\mathrm{1}} \approx\mathrm{29}.\mathrm{46km}\:\mathrm{hr}^{−\mathrm{1}} \right] \\ $$
Commented by Necxx last updated on 09/Oct/18
$${oh}…{Thanks} \\ $$