A plane flies 459 km east from city A to city B in 41.0 min and then 909 km sout

Question

A plane flies 459 km east from city A to city B in 41.0 min and then 909 km south from city B to city C in 1.80 h. For the total trip, what are the (a) magnitude (in km) and (b) direction of the plane's displacement, the (c) magnitude (in km/h) and (d) direction of its average velocity, and (e) its average speed (in km/h)? Give your angles as positive or negative values of magnitude less than 180 degrees, measured from the +x direction (east).

for a)
Magnitude = sqrt(459^2+(-909)^2) = 1018 km
b)
Angle = inverse tangent (-909/459) = -63.2(degrees)
c)
sqrt((459/(41/60))^2 + (-909/1.80)^2)= 840.3
d)Angle = inverse tangent (-505/672)= -36.9

Explanation / Answer

a)

Magnitude = sqrt(459^2+(-909)^2) = 1018 km

b)

Resultant displacement

Direction = inverse tangent (-909/459) = -63.2(degrees)

c)

x- coponent of avg velocity = displacement along x / total time taken

V avg-x = 459 km/2.483 hr =184.8 km/h

y- coponent of avg velocity = displacement along- y / total time taken

Vavg-y = -909km/2.483hr = - 366 km/hr

magnitude of average velocity ={(Vavg-x)2+(Vavg-y)2 }0.5 = 410 km/hr

d)

direction of avg velocity = inverse tangent (-366/184.8) = - 63.2 degrees

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