A cruiser C is sailing due east at a constant speed of 20 kmh-1 and a destroyer D is sailing due North at a constant speed of 10 kmh-1. At noon C and D have position vectors –5i km and –20j km respectively.
a) Show that at time t hours after noon the position vector of C relative to D is given by
[(20t – 5)i + (20 – 10t)j] km

b) Show that the distance d km between the vessels, at this time, is given by
d2 = 25 [20t2 –24t + 17]

c) Hence show that C and D are closest together at 12:36 pm and find
the distance between them at this time.

Question

A cruiser C is sailing due east at a constant speed of 20 kmh-1   and a destroyer D is sailing due North at a constant speed of 10 kmh-1. At noon C and D have position vectors –5i km and –20j km respectively. 
a)	Show that at time t hours after noon the position vector of C relative to D is given by 
			[(20t – 5)i + (20 – 10t)j] km

b)	Show that the distance d km between the vessels, at this time, is given by
				d2 = 25 [20t2 –24t + 17]

c) Hence show that C and D are closest together at 12:36 pm and find    
the distance between them at this time.

anonymous · Answer

I need help with the (c) part only