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A particle's position on the x-axis is given by the function x = t^2 + 6.00 t + 3.00, where t is...

Question:

A particle's position on the x-axis is given by the function {eq}x = t^2 + 6.00 t + 3.00 {/eq}, where {eq}t {/eq} is in second. Where is the particle when {eq}\displaystyle v (x) = 5.00\ \rm m/s {/eq}?

Function:

To find the position of particle, find the derivative of function. After that {eq}v\left ( t \right ) = {x}' \left ( t \right ) {/eq}

With the help of above function, find the value of t.

Now, substitute the value of t in the given function and find the position of particle.

Answer and Explanation:

Given: {eq}x = t^2 + 6.00 t + 3.00 {/eq}

Find the derivative.

{eq}{x}' \left ( t \right ) = 2t + 6.00 {/eq}

Know that, {eq}v\left ( t \right ) = {x}' \left ( t \right ) {/eq}

{eq}2t + 6.00 = 5 {/eq}

{eq}2t = 5 - 6 {/eq}

{eq}2t = -1 {/eq}

{eq}t = - \frac{1}{2} = -0.5 sec {/eq}

When {eq}t = -0.5 {/eq}, then find the value of {eq}x {/eq}.

{eq}x = \left ( -0.5 \right )^2 + 6.00 \left (-0.5 \right ) + 3.00 {/eq}

{eq}= 0.25-3+3 {/eq}

{eq}=0.25 {/eq}

Thus, the value of x is 0.25.


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Basic Calculus: Rules & Formulas

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