A jet is flying due north at 300 m/s with respect to the ground. The wind at its

Question

A jet is flying due north at 300 m/s with respect to the ground. The wind at its altitude is south west and has a speed of 200 m/s. Find the airspeed vector of the plane, and find its position vector as a function of time assuming It is at the origin at time zero. Danielle drives in one direction for five minutes at a constant speed and goes 2400 m. Then she turns right, doubles her speed, and travels for another 200 seconds before stopping. Find her final position, her displacement distance from her starting point, her distance traveled, her average speed, and her average velocity vector.

Explanation / Answer

part 1:

let east be along +ve x axis and north be along +ve y axis.

let i and j are unit vectors along +ve x and +ve y axis respectively.

velocity of the jet=(300 m/s) j

velocity of wind=200*(-cos(45) i - sin(45) j) m/s

then velocity of plane with repsect to wind=velocity of plane with respect to ground-velocity of wind with repsect to ground

=300 j +200*cos(45) i+200*sin(45) j

=141.42 i + 441.42 j m/s

then net speed=sqrt(141.42^2+441.42^2)=463.522 m/s

position vector = speed*time+initial position=(141.42 i+441.42 j)*t meters

part 2:constant speed in the first phase=distancetime=2400/(5*60)=8 m/s

let initially she is travelling along +ve x axis.

if she starts from origin(0,0), then her position after first phase=(2400,0)

then she turns right and moves in -ve y direction

her speed=2*8=16 m/s

she travels for 200 seconds

distance covered=16*200=3200 m

hence her position=(2400,-3200) m

displacement distance=sqrt(2400^2+3200^2)=4000 m

distance travelled=2400+3200=5600 m

average speed=total distance/total time taken=5600/(300+200)=11.2 m/s

average velocity vector=total displacement/total time

=(2400,-3200)/500=(4.8,-6.4) m/s

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