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I have this equation in my book

$x\left(t\right)=x_0+v_0\left(t-t_0\right)+\frac{1}{2}a\left(t-t_0\right)^2$

And it says that it can be written as

$x\left(t\right)=a_1+a_2t+a_3t^2$

Where

$a_1=x_{0\:}-v_0t_0+\frac{1}{2}at_0^2$

$a_2=v_0-at_0$

$a_3=\frac{1}{2}a$

I tried to multiply in and expand everything in parentheses and wound up with

$x\left(t\right)\:=\:x_0+v_0t-v_0t_0+\frac{a}{2}t^2-\frac{a}{2}2tt_0+\frac{a}{2}t_0^2$

So I have a1, but I'm having trouble seeing how they get the other two.

  • 1
    Hint: $x_0$, $t_0$, and $a$ are constants. – 2017-02-14
  • 1
    You just had some extra curly brackets there. Should be clear now. – 2017-02-14

3 Answers 3

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$$x\left(t\right)=x_0+v_0\left(t-t_0\right)+\frac{1}{2}a\left(t-t_0\right)^2$$

$$ x(t) =x_0+v_0t-v_0t_0+\frac{1}{2}a(t^2-2tt_0+t_0^2)$$

$$ x(t)=x_0+v_0t-v_0t_0 + \frac{1}{2}at^2-att_0+\frac{1}{2}at_0^2 $$

$$ x(t)=(x_0-v_0t_0+\frac{1}{2}at_0^2)+(v_0-at_0)t+\left(\frac{1}{2}a \right)t^2$$

$$ x(t)=a_1+a_2t+a_3t^2$$

Where $a_1=x_{0\:}-v_0t_0+\frac{1}{2}at_0^2$

$a_2=v_0-at_0$

$a_3=\frac{1}{2}a$

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After getting $a_1$.You left with,

$=v_0t+\frac{a}{2}t^2-\frac{a}{2}2tt_{0}$

$=v_0t+\frac{a}{2}t^2-att_{0}$

Taking first and last term from above and then take t common,

$=t(v_0-at_{0})$

$=ta_2$

Remaining term,

$\frac{a}{2}t^2$

This is $a_3t^2$

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So the first 3 terms of your expression give you the first two terms of the expression we want, and then you factor the last three terms of your expression to get the last term of what you want!