Asked Sep 13, 2019

A train at a constant 65.0 km/h moves east for 30.0 min, then in a direction 50.0° east of due north for 30.0 min, and then west for 40.0 min. What are the (a) magnitude and (b) angle (relative to east) of its average velocity during this trip?


Expert Answer

Step 1

Given information:


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Mobion 30 miA Te duftorunt 3 Vxt 65x 30 p 32.5 kom The duplaurmnt in eond sgment of motior is S Vxt 32.5Kkm öm druet 50° Caut of Norh T dafphomunt im hind segment of Pobion S-t6Dhh x 40-mwim 6ox 2 pa - 140 Km duų cueut dflomut diagron s follaus 4okm 20-8 Hm Juo Z4.89 32-5 3omim

Step 2

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addia duplormtnds n o douebon Sdu Pa 39.52489 4013.39 1-L Kam 7 daplomnt aony Nordmno)ducbon Sdue nth20 89 m 20.q m 3-49 S = 74 1-t ko-9122 20.a1


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