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A particle moves along the x-axis with its position on the axis at time t given by p(t)=2 t+...

Question:

A particle moves along the x-axis with its position on the axis at time {eq}t {/eq} given by {eq}p(t)=2 t+ \frac{4}{(6-t)} , \ 0<t<6 {/eq}.

At which of the following times is the velocity equal to 3?

Quadratic equation

This type of equation is the second degree and contains at least one squared term and one term with a variable and constant term in these equations. We can find maximum two roots be it real or imaginary.

Answer and Explanation:

Given: {eq}p\left( t \right) = 2t + \dfrac{4}{{6 - t}}\;\;,\,\,0 < t < 6 {/eq}

To find: time at which velocity is 3

Differentiate the given equation:

As differentiation of distance is velocity :

{eq}\begin{align*} p\left( t \right) &= 2t + \dfrac{4}{{6 - t}}\\ \dfrac{{dP\left( t \right)}}{{dt}} &= \dfrac{d}{{dt}}\left( {2t + \dfrac{4}{{6 - t}}} \right)\\ v(t) &= 2 + \dfrac{4}{{{{\left( {6 - t} \right)}^2}}} \end{align*} {/eq}

Value of t when v(t) is 3

{eq}\begin{align*} 3 &= 2 + \dfrac{4}{{{{\left( {6 - t} \right)}^2}}}\\ 3 &= 2 + \dfrac{4}{{36 + {t^2} - 12t}}\\ 3 &= \dfrac{{2\left( {36 + {t^2} - 12t} \right) + 4}}{{36 + {t^2} - 12t}}\\ 3\left( {36 + {t^2} - 12t} \right) &= 72 + 2{t^2} - 24t + 4\\ 108 + 3{t^2} - 36t &= 72 + 2{t^2} - 24t + 4\\ 32 - {t^2} - 12t &= 0\\ {t^2} - 12t + 32 \end{align*} {/eq}

The solution is in form of quadratic equation

So solve the above quadratic equation for t

{eq}\begin{align*} {t^2} - 12t + 32 = 0\\ \left( {t - 4} \right)\left( {t - 8} \right) = 0\\ t = 4,8 \end{align*} {/eq}

So at the value of t at which velocity is 3 are {eq}t=4,8 {/eq}


Learn more about this topic:

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How to Use the Quadratic Formula to Solve a Quadratic Equation

from Math 101: College Algebra

Chapter 4 / Lesson 10
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