Easy Position Vectors question.

Hello,
I am currently stuck on this question.
At 11:00 hours the position vector of an aircraft relative to an airport O is (200i+30j)km, i and j being the unit vectors due east and due north repectively. The velocity of the aircraft is (180i-120j) km h-1
a.) find the time when the aircraft is due east of the airport O
b.) How far it then is from O
c.) how far it is from O at 12:00

Thank you in advance!

Reply 1

Applying4Maths

Original position vector A(200i+30j) at time t=0
Velocity v = (180i-120j)

a) due east of O, position vector B(ki+0j)

B = A + vt
(ki+0j) = (200i+30j) + t(180i-120j)

'j': 0 = 30 - 120t => 120t = 30 => t = 0.25 hr => at 11.15

b) as above, using i

'i': k = 200 + 180t = 200 + 180 x 0.25 = 200 + 45 = 245km

c) t=1, position vector C

(mi+nj) = (200i+30j) + t(180i-120j)

'i': m = 200 + 180t = 200 + 180 x 1 = 200 + 180 = 380
'j': n = 30 - 120t = 30 - 120 x 1 = 30 - 120 = -90

C(380i-90j)

Distance from O = |OC| = sq.root(380^2 + 90^2) = sq.root(144400 + 8100) = sq.root(152500) ? 391km

:smile:

Reply 2

bowers

OP

Applying4Maths

Original position vector A(200i+30j) at time t=0
Velocity v = (180i-120j)

a) due east of O, position vector B(ki+0j)

B = A + vt
(ki+0j) = (200i+30j) + t(180i-120j)

'j': 0 = 30 - 120t => 120t = 30 => t = 0.25 hr => at 11.15

b) as above, using i

'i': k = 200 + 180t = 200 + 180 x 0.25 = 200 + 45 = 245km

c) t=1, position vector C

(mi+nj) = (200i+30j) + t(180i-120j)

'i': m = 200 + 180t = 200 + 180 x 1 = 200 + 180 = 380
'j': n = 30 - 120t = 30 - 120 x 1 = 30 - 120 = -90

C(380i-90j)

Distance from O = |OC| = sq.root(380^2 + 90^2) = sq.root(144400 + 8100) = sq.root(152500) ? 391km

:smile:

Thank you very much!

Easy Position Vectors question.

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