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15 July, 07:47

Shaun is 68 miles away from Keegan. They are traveling towards each other. If Keegan travels 7 mph faster than Shaun and they meet after 4 hours, how fast was each traveling?

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  1. 15 July, 08:13
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    Answer:vs=5mph, vk=7+vs=12mph

    Explanation: kinematics with constant speed

    (shaun) xs = 0 + vs*t

    (keegan) xk = 68 - vk*t

    they meet after t = 4hours (xs=xk):

    vs*t = 68-vk*t

    but we know that vk=7+vs (keegan faster) replacing:

    vs*4 = 68 - (7+vs) * 4

    solving vs=5mph, vk=7+vs=12mph
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