A watermelon seed has the following coordinates: x = -2.8 m, y = 5.3 m, and z =

Question

A watermelon seed has the following coordinates: x = -2.8 m,y = 5.3 m, and z = 0 m. Find its position vector as (a) a magnitude and (b) an angle relative to the positive direction of the x axis. If the seed is moved to the xyz coordinates (8.6 m,0 m,0 m), what is its displacement as (c) a magnitude and (d) an angle relative to the positive direction of the x axis?

hint

(b) and (c) Draw the x and y components in a head-to-tail arrangement in a sketch of the xy plane. The magnitude is the hypotenuse of the right triangle for which those components are the leg. Angle should be reported relative to the positive direction of the x axis, positive if measured counterclockwise and negative if measured clockwise.

(e)-(g) Displacement is a vector that extends from the initial location to the final location. It is calculated by subtracting the initial position vector (the answer to (a)) from the final position vector, both in unit-vector notation.

Explanation / Answer

a) mag = sqrt(2.8^2 + 5.3^2 + 0^2)=5.99 m

b) angle = arctan(y/x) = arctan(5.3/-2.8)=117.85 degrees

c) displacemnet
x = 8.6+2.8
y = -5.3
z = 0

mag = sqrt( (8.6+2.8)^2 + 5.3^2)=12.57

d) angle = arctan( -5.3/(8.6+2.8))=-24.93

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