Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 45085 by Necxx last updated on 08/Oct/18

A motorist travelled from A to B.  This is a distance of 142km at an  average speed of 60kmhr^(−1) .He  spent 5/2hours in B and then  returned to A at an average speed  of 80kmh^(−1) .  a)At what time did the man arrive  back at A  b)find the average speed for the_   total journey.

$${A}\:{motorist}\:{travelled}\:{from}\:{A}\:{to}\:{B}. \\ $$$${This}\:{is}\:{a}\:{distance}\:{of}\:\mathrm{142}{km}\:{at}\:{an} \\ $$$${average}\:{speed}\:{of}\:\mathrm{60}{kmhr}^{−\mathrm{1}} .{He} \\ $$$${spent}\:\mathrm{5}/\mathrm{2}{hours}\:{in}\:{B}\:{and}\:{then} \\ $$$${returned}\:{to}\:{A}\:{at}\:{an}\:{average}\:{speed} \\ $$$${of}\:\mathrm{80}{kmh}^{−\mathrm{1}} . \\ $$$$\left.{a}\right){At}\:{what}\:{time}\:{did}\:{the}\:{man}\:{arrive} \\ $$$${back}\:{at}\:{A} \\ $$$$\left.{b}\right){find}\:{the}\:{average}\:{speed}\:{for}\:{the}_{} \\ $$$${total}\:{journey}. \\ $$

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) ]

$$\mathrm{journey}\:\mathrm{starts}\:\mathrm{at}\:{A}\:\mathrm{at}\:\mathrm{time}=\mathrm{0} \\ $$$$\mathrm{arrival}\:\mathrm{in}\:{B}\:\mathrm{after}\:\frac{\mathrm{142km}}{\mathrm{60km}\:\mathrm{hr}^{−\mathrm{1}} }=\frac{\mathrm{71}}{\mathrm{30}}\mathrm{hr}=\mathrm{2hr22min} \\ $$$$\:\:\:\:\:\mathrm{time}=\mathrm{2hr22min} \\ $$$$\mathrm{5hr30}\:\mathrm{min}\:\mathrm{in}\:{B},\:\mathrm{start}\:\mathrm{from}\:{B}\:\mathrm{at}\:\mathrm{7hr52min} \\ $$$$\mathrm{arrival}\:\mathrm{in}\:{A}\:\mathrm{after}\:\frac{\mathrm{142km}}{\mathrm{80km}\:\mathrm{hr}^{−\mathrm{1}} }=\frac{\mathrm{71}}{\mathrm{40}}\mathrm{hr}=\mathrm{1hr46min30s} \\ $$$$\:\:\:\:\:\mathrm{time}=\mathrm{9hr38min30s} \\ $$$$ \\ $$$$\mathrm{the}\:\mathrm{average}\:\mathrm{speed}\:\left(\mathrm{without}\:\mathrm{the}\:\mathrm{5}:\mathrm{30}\:\mathrm{rest}\right)\:\mathrm{is} \\ $$$$\frac{\mathrm{2}×\mathrm{142km}}{\left(\frac{\mathrm{71}}{\mathrm{30}}+\frac{\mathrm{71}}{\mathrm{40}}\right)\mathrm{hr}}=\frac{\mathrm{480}}{\mathrm{7}}\mathrm{km}\:\mathrm{hr}^{−\mathrm{1}} \approx\mathrm{68}.\mathrm{57km}\:\mathrm{hr}^{−\mathrm{1}} \\ $$$$\:\:\:\:\:\left[\mathrm{with}\:\mathrm{the}\:\mathrm{rest}\:\mathrm{included}\:\mathrm{it}'\mathrm{s}\:\frac{\mathrm{2}×\mathrm{142km}}{\mathrm{9hr38min30s}}=\right. \\ $$$$\left.\:\:\:\:\:\:=\frac{\mathrm{34080}}{\mathrm{1157}}\mathrm{km}\:\mathrm{hr}^{−\mathrm{1}} \approx\mathrm{29}.\mathrm{46km}\:\mathrm{hr}^{−\mathrm{1}} \right] \\ $$

Commented by Necxx last updated on 09/Oct/18

oh...Thanks

$${oh}...{Thanks} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com