Physical applications of calculus question (1 Viewer)

[nazfiz]

Two Straight roads meet at an angle of 60 degrees. Car A starts from the intersection and travels along one road at 40km/h. one hour later Car B starts from the intersection and travels along the other road at 50 km/h. At what rate is the distance between them changing three hours after Car A starts?

This question has got me stumped. Any help would be appreciated

Thanks guys!

 

[nazfiz]

bump

 

[fortune_cookie]

distance = speed x time

Take the reference time t=0 when the second car (car B) starts.

Draw a diagram

Car A will be (40 + 40t)km from the intersection. //in the first hour it travelled 40km , then it travels 40km for ever hour after.

Car B will be (50t)km from the intersection.

Now use cosine rule to find the expression for the distance (y) between them

y^2 = (40+40t)^2 + (50t)^2 - 2(40+40t)(50t)cos(60)

y^2 = 1600(1+t)^2 + 2500t^2 - 40(1+t)(50t)

y^2 = 2100t^2 + 1200t + 1600

Now take the square root of both sides, differentiate and sub t=2 (NOTE: you sub t=2 instead of 3 because you took the reference (t=0) when car B started).

y= (2100t^2 +1200t +1600)^(1/2)


dy/dt = (1/2) (2100t^2 +1200t+1600)^(-1/2) (4200t +1200)

Sub t=2

dy/dt = (1/2) ( 2100(4) +1200(2) +1600)^(-1/2) * (4200(2)+1200) = 43.1km/hr

 

[nazfiz]

distance = speed x time

Take the reference time t=0 when the second car (car B) starts.

Draw a diagram

Car A will be (40 + 40t)km from the intersection. //in the first hour it travelled 40km , then it travels 40km for ever hour after.

Car B will be (50t)km from the intersection.

Now use cosine rule to find the expression for the distance (y) between them

y^2 = (40+40t)^2 + (50t)^2 - 2(40+40t)(50t)cos(60)

y^2 = 1600(1+t)^2 + 2500t^2 - 40(1+t)(50t)

y^2 = 2100t^2 + 1200t + 1600

Now take the square root of both sides, differentiate and sub t=2 (NOTE: you sub t=2 instead of 3 because you took the reference (t=0) when car B started).

y= (2100t^2 +1200t +1600)^(1/2)


dy/dt = (1/2) (2100t^2 +1200t+1600)^(-1/2) (4200t +1200)

Sub t=2

dy/dt = (1/2) ( 2100(4) +1200(2) +1600)^(-1/2) * (4200(2)+1200) = 43.1km/hr

Thanks man! Really helped.