A car drives north at 50 mi/h for 10 min and then turns east and goes 5.0 min at 62 mi/h. Finally, it goes southwest at 26 mi/h for 6.0 min. what is the car's displacement? What is its average velocity for this trip?

1 Answer
Sep 28, 2016

"please have a look the animation"please have a look the animation
v_a=22.06 " "mphva=22.06 mph

Explanation:

enter image source here

bar (AD)=50* 10/60=50/6=8.33 " miles"¯¯¯¯¯¯AD=501060=506=8.33 miles

bar(DE)=62*5/60=31/6=5.17" miles"¯¯¯¯¯¯DE=62560=316=5.17 miles

bar(EB)=26*6/60=156/60=2.6 " miles"¯¯¯¯¯¯EB=26660=15660=2.6 miles

bar(AD)=bar(EC)¯¯¯¯¯¯AD=¯¯¯¯¯¯EC
bar(DE)=bar(AC)¯¯¯¯¯¯DE=¯¯¯¯¯¯AC

"use ABC triangle"use ABC triangle

bar(AC)=bar(DE)=5.17¯¯¯¯¯¯AC=¯¯¯¯¯¯DE=5.17

bar(BC)=bar(EC)-bar(EB)¯¯¯¯¯¯BC=¯¯¯¯¯¯EC¯¯¯¯¯¯EB

bar(BC)=8.33-2.6=5.73¯¯¯¯¯¯BC=8.332.6=5.73

"displacement"=bar(AB)=sqrt(bar(AC)^2+bar(BC)^2)displacement=¯¯¯¯¯¯AB=¯¯¯¯¯¯AC2+¯¯¯¯¯¯BC2

bar(AB)=sqrt((5.17)^2+(5.73)^2)¯¯¯¯¯¯AB=(5.17)2+(5.73)2

bar(AB)=7.72" miles"¯¯¯¯¯¯AB=7.72 miles

"average velocity is described as " v_a=("displacement")/("time elapsed")average velocity is described as va=displacementtime elapsed

"time elapsed="10/60+5/60+6/60=21/60=0.35 " "htime elapsed=1060+560+660=2160=0.35 h

v_a=(bar(AB))/(0.35)va=¯¯¯¯¯¯AB0.35

v_a=(7.72)/(0.35)va=7.720.35

v_a=22.06 " "mphva=22.06 mph