physics---mechanics VECTOR

(a) speed of H = square-root[1.2^2 + 0.9^2] m/s = 1.5 m/s

(b) Position vector of H at time t sec after noon
=[(1.2t)i + (100 - 0.9t)j ]

(c) Position vector of K at t sec after noon
= [(9+0.75t)i + (46+1.8t)j ]

Hence, HK = [(9+0.75t) - 1.2t]i + [(46+1.8t) - (100-0.9t)]j
i.e. HK = [(9 - 0.45t)i + (2.7t - 54)j ]

(d) When H and K meet, HK = 0
i.e. (9 - 0.45t) = 0
t = 9/0.45 s = 20 s

and 2.7t - 54 = 0
t = 54/2.7 s = 20 s
Since both x and y separations become zero after the same time interval of 20 s, this shows H and Y will be at the same position. Hence they will meet.

Position vector of the point wehre they meet
= [(1.2x20)i + (100 - 0.9x20)j]
= 24i + 82j