July 10th, 2015, 07:13 AM  #1 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36  Constant Acceleration.
A cyclist travels from A to B, a distance of 240 metres. He passes A at 12m/s, maintains this speed for as long as he can, and then brakes so that he comes to a stop at B. If the maximum deceleration he can achieve when braking is 3m/s, what is the least time in which he can get from A to B? I've the full solution done my lecturer. But I don't understand it. Can anyone explain this question to me? thanks! 
July 10th, 2015, 09:50 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,837 Thanks: 1479 
velocity vs time graph ... 
July 10th, 2015, 05:18 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,704 Thanks: 669 
I got the same answer, although I don't understand your method. Let s be the time before braking. 12s is the distance before braking. After braking the additional distance traveled is $\displaystyle \int_0^4 (123t)dt=24$. Total distance 12s+24=240, so s = 18 and total time = 22. 
July 11th, 2015, 03:53 AM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,837 Thanks: 1479  

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acceleration, constant 
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