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 May 8th, 2017, 01:21 PM #1 Member   Joined: May 2016 From: Ireland Posts: 96 Thanks: 1 journey a particle moves 13m in the first second and 43m and 83m in the fourth and eight seconds. are the distances consitent with the supposition that it is moving with constant acceleration i am trying to solve this using uvast equations. for each bit i know the time,distance
 May 8th, 2017, 02:38 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 $s(t) = \dfrac {a t^2}{2} + v_0 t + s_0$ the phrasing you've used makes it sound like $s_0=0$ $s(1) = \dfrac{a}{2} + v_0 = 13$ $s(4)=8a+4v_0 = 43$ $s(8) = 32a + 8v_0 = 83$ $\begin{pmatrix}\dfrac 1 2 &1 \\8 &4 \\ 32 &8\end{pmatrix}\begin{pmatrix}a \\ v_0\end{pmatrix} = \begin{pmatrix} 13 \\ 43 \\ 83 \end{pmatrix}$ Ignoring the 3rd row for a moment we solve the system using the first two rows to obtain $a=-\dfrac 3 2,~v_0 =\dfrac{55}{4}$ now we check if the third point fits $83 \overset{?}{=} 32\left(-\dfrac 3 2\right) + 8\left(\dfrac{55}{4}\right)$ $32\left(-\dfrac 3 2\right) + 8\left(\dfrac{55}{4}\right) = 110-48 = 62 \neq 83$ So no these points are not consistent with constant acceleration
 May 9th, 2017, 12:00 AM #3 Member   Joined: May 2016 From: Ireland Posts: 96 Thanks: 1 Thanks very much

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