Question

Particle Motion A particle moves along a line so that its position at any time \(t \geq 0\) is given by the function \(s(t)=\) \((t-2)^{2}(t-4)\) where \(s\) is measured in meters and \(t\) is measured in seconds (a) Find the instantaneous velocity at any time t. (b) Find the acceleration of the particle at any time t. (c) When is the particle at rest? (d) Describe the motion of the particle. At what values of t doesthe particle change directions?

Short answer

The instantaneous velocity function at any time t is \(v(t) = 3t^{2} - 8t + 4\) and the acceleration function is \(a(t) = 6t - 8\). The particle is at rest at times \(t = 4/3\ s\) and \(t = 2\ s\). The particle changes direction at these times as well.

Step-by-step solution

Calculate the derivative of the displacement function

The velocity of the particle at any time \(t\) is the derivative of displacement \(s(t)\). Differentiate \(s(t)\) using the chain rule, resulting in \(v(t) = ds(t)/dt = 3t^{2} - 8t + 4\).

Calculate the derivative of the velocity function

The acceleration at any time \(t\) is the derivative of the velocity \(v(t)\). Differentiate \(v(t)\) resulting in \(a(t) = dv(t)/dt = 6t - 8\).

Solving for when the particle is at rest

The particle is at rest whenever the velocity is zero. To determine when the particle is at rest, set the velocity function \(v(t) = 0\). This results in the equation \(3t^{2} - 8t + 4 = 0\). Solve it for \(t\) by using the quadratic formula and results are \(t = 4/3, 2\). Therefore, the particle is at rest at times \(t = 4/3\ s\) and \(t = 2\ s\).

Describe the motion and find when the particle changes direction

The particle changes direction whenever the velocity changes sign. To determine this, identify the times when the particle is at rest (as found in step 3), and separately consider the intervals of time that they separate. The intervals are \(t < 4/3\), \(4/3 < t < 2\), and \(t > 2\). Substitute a value from each interval into the velocity function, the sign of the result will tell you direction of motion. Results are: For \(t < 4/3\), \(v(t) > 0\), that means the particle moves rightward. For \(4/3 < t < 2\), \(v(t) < 0\), that means the particles moves leftward, and for \(t > 2\), \(v(t) > 0\), again, it moves rightwards. So, the particle changes directions at \(t = 4/3\ s\) and \(t = 2\ s\).