Question

Margot is walking in a straight line from a point 20 feet due east of a statue in a park toward a point 14 feet due north of the statue. She walks at a constant speed of 5 feet per second.(a) Write parametric equations for Margot's position t seconds after she starts walking. (Round your coefficients to four decimal places as needed.)(b) Write an expression for the distance from Margot's position to the statue at time t. (Round your coefficients to four decimal places as needed.)(c) Find the times when Margot is 18 feet from the statue. (Round your answers to two decimal places)t= sec (smaller value)t= sec (larger value)

Answer 1

Answer:

a) x = 20 - 4.0962*t , y = 2.8673*t

b) S = sqrt (25t^2 -163.848t + 400)

c) t = 0.50 s , t = 6.05 s

Step-by-step explanation:

Coordinate system:

Statue position = (0,0)

Initial position  (A) = ( 20 , 0 )

Final position (B) = ( 0 , 14 )

Speed = 5 ft /s

Step 1: Find the velocity vector

Vector (AB) = (0 - 20 ) i  + (14 - 0) j = -20 i + 14 j

mag (AB) = sqrt ( 20^2 + 14^2 ) = 2 sqrt(149)

unit vector (AB) = (-10 / sqrt(149) ) i + ( 7 / sqrt(149) ) j

Velocity (AB) = 5*(-10 / sqrt(149) ) i + 5*( 7 / sqrt(149) ) j

Velocity (AB) = (-50 / sqrt(149) ) i + ( 35 / sqrt(149) ) j

Step 2: Find parametric equation for both x and t

x = x(0) + V_ab . i*t

x = 20 - 50 / sqrt(149) * t

y = y(0) + V_ab . j*t

y = 35 / sqrt(149) * t

Hence,

x = 20 - 4.0962*t

y = 2.8673*t

part b

S = sqrt ( x^2 + y^2 )

S = sqrt ( (20 - 4.0962*t)^2 + (2.8673*t)^2 )

S = sqrt (25t^2 -163.848t + 400)

part c

S = 18 ft

sqrt (25t^2 -163.848t + 400) = 18

25t^2 -163.848t + 76 = 0

Solve for t:

t = 0.50 s

t = 6.05 s