A train starting at 11 am travels east at 45 km / hr while another starting at 12 noon from the same point trivels south at 60 km / hr . how fast the train separating at 3 p. m

Answer 1

train 1, x= 45*3 = 135, y = 0
dx/dt = 45

train 2, x = 0, y= -60*3 = -180
dy/dt = -60

z = hypotenuse
z^2 = x^2+y^2 = 3^2 (45^2+60^2)
= 3^2 (2025 + 3600) = 3^2(5625)
so
z = 3*75 = 225

z^2 = x^2 + y^2
2 z dz/dt = 2 x dx/dt +2 y dy/dt

dz/dt = (xdx/t+ydy/dt)/z
= (135*45-180*-60)/225
=etc

Answer 2

at time t hr after second train travels from starting point

train 1 , X = 45 * (t+1)
train 2 , Y = 60 * t
distance apart

s^2 = [ 60^2 t^2 + 45^2 ( t+1)^2]

2s ds/dt = 2 * 3600* t + 2025 * 2 (t+1)

2s ds/dt = 7200 t + 4050 t + 4050

2s ds/dt = 11250 t + 4050

s ds/dt = 5625 t + 2025

when t= 3
s2= 32400+32400= 64800
s=254.56

ds/dt = 18900/ 254.56
= 74.245 mph ------answer