An autonomous drone navigating an obstacle course experiences three mutually perpendicular… — JAMB UTME

An autonomous drone navigating an obstacle course experiences three mutually perpendicular velocity vectors simultaneously: a horizontal forward velocity of $12\text{ m s}^{-1}$ due East, a horizontal crosswind drift velocity of $9\text{ m s}^{-1}$ due North, and a vertical downdraft velocity of $8\text{ m s}^{-1}$ directed straight downward. What is the magnitude of the resultant velocity vector of the drone in $\text{m s}^{-1}$ ?

Answer: 17 m s^-1

Answer

17 m s⁻¹

Because the three velocity components are mutually perpendicular, the magnitude of the overall resultant velocity is calculated using the 3D Pythagorean theorem:

\sqrt{12^2 + 9^2 + 8^2} = \sqrt{144 + 81 + 64} = \sqrt{289} = 17\text{ m s}^{-1}

Step-by-Step Solution

Identify the perpendicular vector components

v_x = 12\text{ m s}^{-1}

v_y = 9\text{ m s}^{-1}

v_z = 8\text{ m s}^{-1}

The three given velocity vectors act along mutually orthogonal spatial axes (East, North, and Downward).

Apply the 3D vector resultant magnitude formula

v_r = \sqrt{v_x^2 + v_y^2 + v_z^2}

Since the vector components are perpendicular to one another, the magnitude of their resultant is given by the extension of the Pythagorean theorem to three dimensions.

Substitute values and evaluate

v_r = \sqrt{144 + 81 + 64} = \sqrt{289} = 17\text{ m s}^{-1}

Squaring each component, adding them, and taking the principal square root yields the total magnitude of the velocity vector.

Key Concept

Magnitude of three mutually perpendicular vector components using the 3D Pythagorean theorem

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