A particle moves along a straight line OX. At a time t (in seconds), the displacement x (in metres) of the par

Question

A particle moves along a straight line OX. At a time t (in seconds), the displacement x (in metres) of the particle from O is given by x = 40 + 12t - t³. How long would the particle travel before coming to rest?

Accepted Answer

1.  **Find Velocity ($v$):** Velocity is the rate of change of position with respect to time ($v = dx/dt$) .
    Given $x = 40 + 12t - t^3$.
    $$ v = \frac{d}{dt}(40 + 12t - t^3) = 12 - 3t^2 $$
2.  **Determine when the particle rests:** The particle comes to rest when its velocity is zero.
    $$ 12 - 3t^2 = 0 $$
    $$ 3t^2 = 12 \implies t^2 = 4 \implies t = 2 	ext{ s} $$
3.  **Calculate Positions:**
    *   Initial position at $t = 0$: $x(0) = 40 + 12(0) - 0^3 = 40$ m.
    *   Position at rest $t = 2$: $x(2) = 40 + 12(2) - 2^3 = 40 + 24 - 8 = 56$ m.
4.  **Calculate Distance:** Since the particle moves in a straight line without reversing direction until $t=2$ (velocity is positive for $t < 2$), the distance travelled is the magnitude of the displacement.
    $$ 	ext{Distance} = |x(2) - x(0)| = |56 - 40| = 16 	ext{ m} $$

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