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