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Kinematics of a Particle in a Plane

Example: Sally is walking east along the x-axis. Her position vector at any time t is r_S =40t i ft.

John is walking south. His position vector at any time t is r_J = 40 i ˗ 20t j ft. See the

figure below. Find the minimum distance between Sally and John and the time that it occurs.

Solution: First determine the position vector from John to Sally. Then the distance, D, between

John and Sally at any time t is the magnitude of this vector. Take the derivative with respect

to time and set it to zero to find the maximum or minimum distance. Use this value of time

to determine the minimum distance.

r_J + r_JS = r_S So r_JS = r_S ˗ r_J

r_JS = 40t i ˗ (40 i ˗ 20t j) = ( 40t ˗ 40) i + 20t j

D = √ [( 40t ˗ 40)² + ( 20t)² ] Now set dD/dt = 0

dD/dt = (1/2) [( 40t ˗ 40)² + ( 20t)² ]^˗1/2[ 2(40t ˗ 40)(40) + 2(20t)(20) ] = 0

80t ˗ 80 + 20t = 0 100t = 80 t = 0.8 sec (result)

Put this value of t into D. D = √ [ (32 ˗ 40)² + (16)² ] = 17.9 ft (result)

Note: When t = 0 D = 40 ft and John is 40 ft to the right of Sally.