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October 22nd, 2019, 01:50 PM   #1
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Math Focus: Calculus
Question 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?
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October 22nd, 2019, 02:51 PM   #2
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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?
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