# 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?

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.

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}