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 October 22nd, 2019, 01:50 PM #1 Senior Member   Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus How do I find the distance traveled by an object during a specified second? The problem is as follows: An object starts moving from rest and accelerates. On which second the distance traveled by such object will be equal to the triple of the distance traveled by that object on the fifth second?The alternatives given on my book are: $\begin{array}{ll} 1.&\textrm{12 th}\\ 2.&\textrm{11 th}\\ 3.&\textrm{13 th}\\ 4.&\textrm{15 th}\\ 5.&\textrm{14 th} \end{array}$ Okay what I attemped to do in this problem was to refer to the equation for accelerated motion as shown below: $x(t)=x_{o}+v_{0x}t+\frac{1}{2}at^2$ As it mentions that starts from rest then the initial condition would be $v_{ox}=0$ ; Now the second part I'm assuming that starts from $x_{o}=0$ which in other words would be right from the start of the reference point. This would make the equation like this: $x(t)=\frac{1}{2}at^2$ Since it mentions the distance traveled to the $5^{\textrm{th}}$ this would be plugin that number in the above equation to find the distance traveled to that point. $x(5)=\frac{1}{2}a\left(5\right)^2=\frac{25}{2}a$ Because it mentions what second would the distance be triple of that I assumed equating the first equation with an unknown time to three times the third equation would render the result. $\frac{3\times 25}{2}a = \frac{1}{2}at^2$ Therefore: $3 \times 25 = t^2$ $t= 5 \sqrt 3$ But oddly this doesn't appears within the alternatives. For this part I'm confused. What exactly did I mistakenly understood? Can somebody help me with the procedure I took?. I revised the way how I did the steps algebraically and it looks okay, but I'm still confused if what I attempted to do was not right?. Can somebody guide me in the right path?  October 22nd, 2019, 02:51 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677 distance traveled in the 5th second is $x(5)-x(4)$ $v_0 = 0$ $x(5) = \dfrac{1}{2}a \cdot 5^2 = \dfrac{25a}{2}$ $x(4) = \dfrac{1}{2}a \cdot 4^2 = 8a$ $x(5)-x(4) = \dfrac{9a}{2}$ want $x(t) - x(t-1) = \dfrac{27a}{2}$ $x(t) = \dfrac{1}{2}a \cdot t^2 = \dfrac{at^2}{2}$ $x(t-1) = \dfrac{1}{2}a \cdot (t-1)^2 = \dfrac{a(t-1)^2}{2}$ $\dfrac{at^2}{2} - \dfrac{a(t-1)^2}{2} = \dfrac{27a}{2}$ $t^2 - (t-1)^2 = 27$ can you finish? Thanks from topsquark Tags distance, find, object, traveled Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post stacigurl12 Calculus 5 February 12th, 2014 12:55 AM mwhit Calculus 6 September 29th, 2013 06:12 AM lovetolearn Calculus 5 April 15th, 2012 02:07 PM justinh8 Algebra 3 March 27th, 2012 02:29 PM Recipe Algebra 3 March 16th, 2010 07:43 PM

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