Hard M1 Stuff - please help

AT 12 noon the position vectors r and the velocity vectors v of two ships A and B are

rA = (2i + j) Km vA = (3i + j) Kmh^-1
rB = (-i -4j) Km vB = (11i + 3j) Kmh^-1

a) Show that at time t h after noon the position vector of B relative to A is given by

[(8t-3)i + (2t-5)j] Km

b) Show that the distance d Km between the vessels, at this time, is given by

d^2 = 68[t^2 - t + 1/2]

c) Hence show that the ships are nearest together at 12.30 p.m. and find the distance between them at this time.

So the position of each is given by its starting position plus its velocity*time

Position A = (2i + j + 3ti + tj)
Position B = (-i - 4j + 11ti + 3tj)

The position of B relative to A is Position B - Position A
= (-i - 4j + 11ti + 3tj) - (2i + j + 3ti + tj)
= -3i - 5j + 8ti + 2tj
= [(8t-3)i + (2t-5)j]



Just use Pythagorus, using i and j as side lengths, thus:
Distance² = (8t-3)² + (2t-5)²
Distance² = 64t² - 48t + 9 + 4t² - 20t + 25
Distance² = 68t² -68t + 34
Distance² = 68(t² - t + ½)



Now we have an equation we can differentiate for highest time, so:

f(x) = 68t² - 68t + 34
f'(x) = 136t - 68
136t - 68 = 0
t = ½

So closest position is after ½ an hour, so at 12.30

Then insert back into equation for the distance, not forgetting to square root.

By the way thanks all of you for your posts - all have been helpful

mm thats the thing i havent done differentiation yet and when i tried to teach myself i didnt understand it - anyone know any good sites?

You don't have to use differentiation, Steve's method worked just as well.

oh yes after seeing steves post i remembered that minimum point can be achieved by completing the square but i was just wondering as an off topic any good sites for differentiation as it will be useful for my C1 exam

My tutor taught us another method...mush easier if you're in hurry. It always works.

minimum point= -b/2a

Try www.s-cool.co.uk ...looks very childish but explains things very clearly.