My Math Forum journey

 Applied Math Applied Math Forum

 May 8th, 2017, 12:21 PM #1 Member   Joined: May 2016 From: Ireland Posts: 94 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, 01:38 PM #2 Senior Member     Joined: Sep 2015 From: CA Posts: 1,264 Thanks: 650 $s(t) = \dfrac {a t^2}{2} + v_0 t + s_0$ 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 8th, 2017, 11:00 PM #3 Member   Joined: May 2016 From: Ireland Posts: 94 Thanks: 1 Thanks very much

 Tags journey

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post InkSprite New Users 1 December 30th, 2016 03:41 PM markosheehan Applied Math 4 December 16th, 2016 02:05 PM russphelan Calculus 26 October 17th, 2016 11:14 AM ThomasAnderson New Users 4 January 25th, 2015 09:23 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top