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 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: 2,924 Thanks: 1521 velocity vs time graph ...
 July 10th, 2015, 05:18 PM #3 Global Moderator   Joined: May 2007 Posts: 6,756 Thanks: 696 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 $\displaystyle \int_0^4 (12-3t)dt=24$. Total distance 12s+24=240, so s = 18 and total time = 22.
July 11th, 2015, 03:53 AM   #4
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Quote:
 Originally Posted by mathman I got the same answer, although I don't understand your method.
Area under the velocity graph is displacement.

Area under the velocity graph is a trapezoid ...

$h=12$, $b_1=t$, $b_2=t+4$

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