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We proved in first part that ABP is congruent to BCQ.
Yes.
Ratio Method
(i) The velocity vector is in the form (xi +yj).
thnk u very much(i) The velocity vector is in the form (xi +yj).
The magnitude is given (10√2), this means that 10√2 = √x^2 +y^2.
The direction is north - east, this means that a angle of 45 deg. is subtended with both y-axis and x-axis (see diagram).
For calculating the component vectors, take trigonometric ratios.
cos 45 = x/10√2
x= 10
cos 45 = y/10√2
y = 10
Therefore, the velocity vector is (10i + 10j).
(ii) Position vector = Initial position vector + distance travelled.
= (-4i+8j) + 2(10i+10j) = 16i +28j.
(iii) V(p) = V(p/q) + V(q) [V(p) = Velocity of P, V(q) = Velocity of Q, V(p/q) = Velocity of P relative to Q].
V(p/q) = V(p) - V(q)
V(p/q) = (10i + 10j) - (8i + 6j) = 2i + 4j
(iv) When they meet, Position vector of P = Position vector of Q,
Hence,
(16i +28j) + t(10i+10j) = (19i + 34j) + t(8i + 6j) (Initial vector at 12 00 + Distance traveled in t hours).
Comparing coefficients of i,
16+10t = 19 +8t
2t = 3
t = 1.5 hours.
Time at which they meet = 12 00 + 1.30 = 13 30 hours.
Position vector at that time (take either of the two) = (16i + 28j) + 1.5(10i + 10j)
Position vector at that time (take either of the two) = 16i + 28j + 15i +15j = 31i + 43j.
Hope that helps!
View attachment 24140
No problem at all.thnk u very much
Yes you will.
Split the inequalities,
The marking scheme is correct,
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