Question

Two parallel paths 17 m apart run east–west through the woods. Brooke walks east on one path at 4 km/h, while Jamail walks west on the other path at 7 km/h. If they pass each other at time ????=0, how far apart are they 11 s later, and how fast is the distance between them changing at that moment?

Answer 1

D(11) = 37.660 m

dD/dt = 2.7260 m/s

Step-by-step explanation:

given data

two path apart = 17 m

walks east one path = 4 km/h = 1.111 m/s

walks west other path = 7 km/h = 1.944 m/s

pass each other time t = 0

solution

we consider here east is the positive direction and west is the negative direction

so that

the east - west distance between them is = 1.111 + 1.944 = 3.055 m/s

and

the actual distance between them time t is

D(t) =
`√((3.055 t)^2 + 17 ^2)`

at time 11 s

D(11) =
`√((3.055 *11)^2 + 17 ^2)`

D(11) = 37.660 m

and

increase rate is dD/dt

dD/dt =
`(0.5(2) t (3.055)^2)/(√((3.055 t)^2 +17^2))`

so for 11 sec

dD/dt =
`(0.5(2) 11 (3.055)^2)/(√((3.055 *11)^2 +17^2))`

dD/dt = 2.7260 m/s