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Suppose we're given the position \$x\$ , velocity \$\dot{x}\$ , and acceleration \$\ddot{x}\$ at \$t_1 \$.

We need to approximate the position \$x\$ at \$t_2\$.

In below picture, I understand how \$x_1\$ represents the vertical black length, and how \$ \color{blue}{\dot{x_1}(t_1)}h\$ represents the vertical blue length. However I don't get how \$\color{green}{\ddot{x}(t_1)}\dfrac{h^2}{2}\$ represents the green length. I'm wondering if there a simple derivation for proving that the green vertical length equals \$\color{green}{\ddot{x}(t_1)}\dfrac{h^2}{2}\$ . Any ideas ? Thanks ! enter image description here

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  • \$\begingroup\$ This question is from signals and systems, but I completely understand if this question belongs in math and I'll gladly post it in math if you think so.. (I got so used to electronics page that I didn't realize this till after finishing composing the q..) \$\endgroup\$
    – AgentS
    Commented Jun 15, 2018 at 7:40
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    \$\begingroup\$ I think this is indeed better suited to Math SE. I'm usually not too strict on this type of stuff, but I believe the connection to electronics is too weak for this question ;-) \$\endgroup\$
    – Sven B
    Commented Jun 15, 2018 at 8:14
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    \$\begingroup\$ Integral of h is.......? \$\endgroup\$
    – Andy aka
    Commented Jun 15, 2018 at 8:44
  • \$\begingroup\$ @Andyaka Integral of h is h^2/2. That quadratic is what we get by integrating x''(t) = c. I now see clearly how the taylor quadratic looks geometrically. Thank you so much :) \$\endgroup\$
    – AgentS
    Commented Jun 15, 2018 at 9:09
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    \$\begingroup\$ Yes @LongPham if I recall correctly, physicists use \$\dot{x}\$ to represent derivative of \$x\$ with respect to time, and mathematicians use \$x'\$... Not sure about engineers hmm \$\endgroup\$
    – AgentS
    Commented Jun 15, 2018 at 9:59

1 Answer 1

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It is the contribution of the constant acceleration a.

After h time, the speed will have increased by ah. So the average extra speed over h will be ah/2. The extra distance due to that average extra speed will then be h*(ah)/2.

Which is the same as

$$ ah^2/2 $$

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  • \$\begingroup\$ Ah so \$ah^2/2\$ is \$a(t_2-t_1)^2/2\$, and this term is part of the whole quadratic \$x(t_1) + x'(t_1)(t_2-t_1) + x''(t_1)(t_2-t_1)^2/2\$ . I see... that average acceleration interpretation is clever! Can we interpret something similar for the fourth term too ? \$ah^3/6\$ \$\endgroup\$
    – AgentS
    Commented Jun 15, 2018 at 9:10
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    \$\begingroup\$ Yeah, it's called jerk \$\endgroup\$ Commented Jun 15, 2018 at 10:21

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