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October 14th, 2016, 08:44 AM   #11
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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?
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Last edited by Benit13; October 14th, 2016 at 08:49 AM.
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October 14th, 2016, 11:18 AM   #12
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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.
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