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 ) write an expression for the distance from margot's position to the statue at time t. (round your coefficients to four decimal places as ) 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)

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

yeet

step-by-step explanation:

step-by-step explanation:

∆abc has the points a(1, 7), b(-2, 2), and c(4, 2) as its vertices. formed with the point d(1, 2) as its third vertex, then ∆abd is triangle.