October 14th, 2016, 08:44 AM  #11 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,142 Thanks: 726 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
Much better, but still not perfect... Firstly... the acceleration at t=4 is the same as the acceleration at t = 3. It's also the same at point t=4.5, t=4.9, t=4.999.... so... basically, the acceleration shouldn't change until t=5, at which point the acceleration jumps to zero. Secondly... regarding the final part of the curve. Consider the following situation. Take an object like a pencil or ruler and drop it. As soon as you let go, it will start to accelerate at a constant rate under gravity. Because it's accelerating at a constant rate, the velocity increases in a linear fashion. If you look at the position, the position increases like a parabolic curve (because the distance it can cover is increasingly very rapidly, like the $\displaystyle y = x^2$ curve). So you have something like: acceleration > constant velocity > linear position > $\displaystyle x^2$ curve (ish) There's a similar pattern for constant velocity: acceleration > 0 velocity > constant position > linear See? there's a pattern. As you consider position to velocity to acceleration, you have 0 > constant > linear > $\displaystyle x^2$ curve. So now... if the velocity is a $\displaystyle x^2$like curve, what do you think the acceleration is going to be like? Last edited by Benit13; October 14th, 2016 at 08:49 AM. 
October 14th, 2016, 11:18 AM  #12 
Global Moderator Joined: Dec 2006 Posts: 20,375 Thanks: 2010 
The velocity curve for A when t > 8 doesn't look parabolic to me, though there is some similarity to a parabola. Its slope for t = 10 is hard to estimate accurately, but appears to be about 9m/s². Vehicle A's acceleration vs time graph for 8 < t < 11 seems to be a gentle curve rather than a straight line. 

Tags 
acceleration, distance, kinematics, question, velocity 
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