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-t

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

$$AmotoristtravelledfromAtoB.$$$$Thisisadistanceof142kmatan$$$$averagespeedof60{kmhr}^{-1}.He$$$$spent5/2hoursinBandthen$$$$returnedtoAatanaveragespeed$$$$of80{kmh}^{-1}.$$$$a)Atwhattimedidthemanarrive$$$$backatA$$$$b)findtheaveragespeedfor{the}_{}$$$$totaljourney.$$

Answered by MJS last updated on 08/Oct/18

$$\mathrm{journey}\mathrm{starts}\mathrm{at}A\mathrm{at}\mathrm{time}=0$$$$\mathrm{arrival}\mathrm{in}B\mathrm{after}\frac{142\mathrm{km}}{60\mathrm{km}{\mathrm{hr}}^{-1}}=\frac{71}{30}\mathrm{hr}=2\mathrm{hr}22\min$$$$\mathrm{time}=2\mathrm{hr}22\min$$$$5\mathrm{hr}30\min \mathrm{in}B,\mathrm{start}\mathrm{from}B\mathrm{at}7\mathrm{hr}52\min$$$$\mathrm{arrival}\mathrm{in}A\mathrm{after}\frac{142\mathrm{km}}{80\mathrm{km}{\mathrm{hr}}^{-1}}=\frac{71}{40}\mathrm{hr}=1\mathrm{hr}46\min 30\mathrm{s}$$$$\mathrm{time}=9\mathrm{hr}38\min 30\mathrm{s}$$$$$$$$\mathrm{the}\mathrm{average}\mathrm{speed}(\mathrm{without}\mathrm{the}5:30\mathrm{rest})\mathrm{is}$$$$\frac{2\times 142\mathrm{km}}{(\frac{71}{30}+\frac{71}{40})\mathrm{hr}}=\frac{480}{7}\mathrm{km}{\mathrm{hr}}^{-1}\approx 68.57\mathrm{km}{\mathrm{hr}}^{-1}$$$$[\mathrm{with}\mathrm{the}\mathrm{rest}\mathrm{included}{\mathrm{it}}^{\prime }\mathrm{s}\frac{2\times 142\mathrm{km}}{9\mathrm{hr}38\min 30\mathrm{s}}=$$$$=\frac{34080}{1157}\mathrm{km}{\mathrm{hr}}^{-1}\approx 29.46\mathrm{km}{\mathrm{hr}}^{-1}]$$

Commented by Necxx last updated on 09/Oct/18

$$oh\dots Thanks$$

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