Question

You are standing on the ground at the origin of a coordinate system. An airplane files over you with constant velocity parallel to the x axis at a fixed height of 7.60x10³m. At time t = 0, the airline is directly above you so that the vector leading from you to it is P₀ = 7.60x10³jm. At t = 30.0s , the position vector leading from you to the airplane is P₃₀ = (8.04x10³i + 7.60x10³j)m as suggested . Determine the magnitude and orientation of the airplane’s position vector at t = 45.0s.

Short answer

The magnitude is 1.43x10⁴ m and the direction is 32.2^o above the horizontal.

Step-by-step solution

Definition of vector

A vector is a quantity that has both a direction and a magnitude, and is used to determine the relative location of two points in space.

Formula

If vector A = a₁i + a₂j + a₃k and vector B = b₁i + b₂j + b₃k then the displacement of AB = (b₁ - a₁)i + (b₂ - a₂)j + (b₃ - a₃)k

Given Parameters

The y coordinate of the airplane is constant and equal to 7.60x10³ m whereas the x coordinate is given by x = v₁t where v₁ is the constant speed in the horizontal direction

When t = 30.0 s m, so$v_1=\frac{8.04\times {10}^3}{30}=268m/s$

So, position vector $P=\left(268\right)i+\left(7.60\times {10}^3\right)j$

When t = 45.0 s , position vector P = (1.21x10⁴),

Calculation

So, the magnitude is

$P=\sqrt{{\left(1.21\times {10}^4\right)}^2+{\left(7.60\times {10}^3\right)}^2}=1.43\times {10}^{4}m$

And the direction is

$\theta ={\tan }^{-1}\left(\frac{7.60\times {10}^3}{1.21\times {10}^4}\right)=32.2^o$above the horizontal .