A bicycle travels 3.2 km due east in 0.10 h, then 4.0 km at 15.0° east of north

Question

A bicycle travels 3.2 km due east in 0.10 h, then 4.0 km at 15.0° east of north in 0.30 h, and finally another 3.2 km due east in 0.10 hto reach its destination. The time lost in turning is negligible.What is the average velocity for the entire trip?
Magnitude___km/h Direction: ___ degrees north of east A bicycle travels 3.2 km due east in 0.10 h, then 4.0 km at 15.0° east of north in 0.30 h, and finally another 3.2 km due east in 0.10 hto reach its destination. The time lost in turning is negligible.What is the average velocity for the entire trip?
Magnitude___km/h Direction: ___ degrees north of east

Explanation / Answer

find the resultant for his trip: Rx=3.2cos0+4cos(90-15)+3.2cos0=7.44km east Ry=3.2sin0+4sin(90-15)+3.2sin0=3.86km north so his total displacement is: |R|=(Rx2+Ry2)=8.38km and the total time is: .1+.3+.1=.5hrs so the average velocity Vavg=x/t=8.38/.5=16.76km/hr use trigonometry to find the direction: =tan-1(Ry/Rx)=27.42o northof east

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