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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?

Respuesta :

encali2008

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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