The position of a particle with respect to time t along the x-axis is given by x = 9t² - t³ where x is in metr

Question

The position of a particle with respect to time t along the x-axis is given by x = 9t² - t³ where x is in metres and t in seconds. What will be the position of this particle when it achieves maximum speed along the +x-direction?

Accepted Answer

1.  **Find Velocity ($v$):** Velocity is the first derivative of position with respect to time ($v = dx/dt$) .
    $$ v = \frac{d}{dt}(9t^2 - t^3) = 18t - 3t^2 $$
2.  **Condition for Maximum Speed:** The speed is maximum when the acceleration ($a = dv/dt$) is zero .
    $$ a = \frac{dv}{dt} = \frac{d}{dt}(18t - 3t^2) = 18 - 6t $$
    Set $a = 0$ to find the critical time:
    $$ 18 - 6t = 0 \implies t = 3 	ext{ s} $$
3.  **Calculate Position:** Substitute $t = 3$ s into the original position equation.
    $$ x = 9(3)^2 - (3)^3 $$
    $$ x = 9(9) - 27 $$
    $$ x = 81 - 27 = 54 	ext{ m} $$

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