A fugitive tries to hop on a freight train traveling at aconstant speed of 5.0 m

Question

A fugitive tries to hop on a freight train traveling at aconstant speed of 5.0 m/s. Just as an empty box car passeshim, the fugitive starts from rest and accelerates at a = 1.2m/s2 to his maximum speed of 6.0 m/s. (a) How long doesit take him to catch up to the empty box car? (b) What is the distance traveled to reach the box car? A fugitive tries to hop on a freight train traveling at aconstant speed of 5.0 m/s. Just as an empty box car passeshim, the fugitive starts from rest and accelerates at a = 1.2m/s2 to his maximum speed of 6.0 m/s. (a) How long doesit take him to catch up to the empty box car? (b) What is the distance traveled to reach the box car?

Explanation / Answer

a) let the time be t, we have: for the box car distance is : d2 = vt (where v =5.0m/s) since max speed is u = 6.0m/s and takes t2 to reach: u = 6.0m/s = 1.2m/s2*t2 =>t2 = 5s distance to catch up d1 = (1/2)a (t2)2  + u*(t-t2) =(1/2)*1.2m/s2*(5s)2 + 6.0m/s*(t-5)               =15 m + 6.0t - 30.0m = -15.0m+6.0t d1 = d2 => -15.0m + 6.0t = vt = 5.0t =>t = 15.0s, it take 15.0s to catch up b) the distance is d2 = 5.0m/s*15.0s = 75.0m for the box car distance is : d2 = vt (where v =5.0m/s) since max speed is u = 6.0m/s and takes t2 to reach: u = 6.0m/s = 1.2m/s2*t2 =>t2 = 5s distance to catch up d1 = (1/2)a (t2)2  + u*(t-t2) =(1/2)*1.2m/s2*(5s)2 + 6.0m/s*(t-5)               =15 m + 6.0t - 30.0m = -15.0m+6.0t d1 = d2 => -15.0m + 6.0t = vt = 5.0t =>t = 15.0s, it take 15.0s to catch up b) the distance is d2 = 5.0m/s*15.0s = 75.0m

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