Question

The function \(s(t)\) describes the motion of a particle moving along a line. For each function, (a) find the velocity function of the particle at any time \(t \geq 0\), (b) identify the time interval(s) when the particle is moving in a positive direction, (c) identify the time interval(s) when the particle is moving in a negative direction, and (d) identify the time(s) when the particle changes its direction. $$ s(t)=6 t-t^{2} $$

Short answer

The velocity function is \(v(t) = 6 - 2t\). The particle is moving in a positive direction for \(t < 3\), moving in a negative direction for \(t > 3\), and changes its direction at \(t = 3\).

Step-by-step solution

Determine the velocity function

To determine the velocity, differentiate the function \(s(t) = 6t - t^{2}\) with respect to \(t\). This gives \(v(t) = 6 - 2t\).

Identify when the particle is moving in a positive direction

The particle is moving in a positive direction when the velocity is greater than zero. So solve the inequality \(v(t) > 0\), which gives \(6 - 2t > 0\). Solving this inequality yields \(t < 3\). So, the particle is moving in the positive direction for \(t < 3\).

Identify when the particle is moving in a negative direction

The particle is moving in a negative direction when the velocity is less than zero. So solve the inequality \(v(t) < 0\), which gives \(6 - 2t < 0\). Solving this inequality yields \(t > 3\). So, the particle is moving in a negative direction for \(t > 3\).

Identify when the particle changes its direction

The particle changes its direction when the velocity changes sign. This occurs at the roots of the velocity function, \(v(t) = 6 - 2t = 0\). Solving this equation yields \(t = 3\). So, the particle changes its direction at \(t = 3\).