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Let s_{a} and s_{b} be the distance of car A and B respectively

v_{a} = s_{a}/t_{a}

s_{a} = v_{a}t_{a}

s_{b} = v_{b}t_{b}

I) the distance between both cars = s_{a} - s_{b}

= v_{a}t_{a} - v_{b}t_{b} ---------- eqn(1)

Assuming 6:40 am as the reference time,

for car A after 1 hour (3600 seconds),

t_{a} = 3600 - 0 = 3600 seconds

For car B, which is 40 minutes (2400 seconds) ahead

t_{b }= 3600 + 2400 = 6000 seconds

from eqn(1),

= 280(3600) - 80(6000)

= 528,000 m

= 528km

Ii) car A overtakes car B when s_{a} = s_{b}

v_{a}t_{a} = v_{b}t_{b}

t_{a} = t

t_{b} = t + 40

280t = 80(t + 40)

280t = 80t + 3200

280t - 80t = 3200

200t = 3200

t = 3200/200

t = 16 minutes (or 960 seconds)

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