Tam travels to class, which is 500 meters from Mats location. On reaching class,

Question

Tam travels to class, which is 500 meters from Mats location. On reaching class, Tam realizes she left her backpack with Mat, and she begins to head directly back at a uniform velocity of 2.0 m/s. At exactly the same time, Mat recives a text message from Tam, and starts from rest but accelerates at a uniform rate of 1.0 m/s2 directly toward Tam in order to bring Tam her backpack sooner. How much time elapses from the moment Tam doubles back and Mat leaves to meet her to the point at which they meet? How far from Mat's initial position do they meet?

Explanation / Answer

distance = 500 m

velocity of tam towards mat is V1 = 2m/s constant

velocity of mat towards tam is V2 = 0 + 1*t = t

distance covered by tam = 2t

distance by mat = 0.5*t^2

total distance covered by both = 0.5t^2 + 2t = 500

=> t^2 + 4t - 1000 = 0

t1 = 29.69 s and t2 = - 33.7 (not possible).

time elapsed = 29.69s

b) distance by mat = 0.5*t^2 = 0.5*29.69^2 = 440.74 m

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