Example 16 A particle moves along a straight line such that its displacement \( s \) (in metres) from a fixed point O is given by \( s=8 t^{3}-15 t^{2}+6 t \), where \( t \) is the time in seconds after passing \( O \). It is given velocity \( v=\frac{d s}{d t} \) and acceleration \( a=\frac{d v}{d t}=\frac{d^{2} s}{d t^{2}} \). \( \rightarrow v=0 \) (a) Find the time when the particle is instantaneously at rest. (b) Find the acceleration after 4 seconds. a) Solve for \( t \) when \( v=0 \) \[ v=\frac{d s}{d t}=24 t^{2}-30 t+6 \]

To find the time when the particle is instantaneously at rest, we need to solve for \( t \) when \( v=0 \). Setting the velocity equation to zero gives us: \[ 24 t^{2} - 30 t + 6 = 0 \] Now we'll use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 24, b = -30, c = 6 \). Calculating the discriminant: \[ b^2 - 4ac = (-30)^2 - 4 \times 24 \times 6 = 900 - 576 = 324 \] Now plugging values into the quadratic formula: \[ t = \frac{30 \pm \sqrt{324}}{2 \times 24} = \frac{30 \pm 18}{48} \] This gives us: \[ t = \frac{48}{48} = 1 \quad \text{and} \quad t = \frac{12}{48} = \frac{1}{4} \] Thus, the particle is instantaneously at rest at \( t = 1 \) second and \( t = \frac{1}{4} \) seconds. Next, for the acceleration after 4 seconds, we first find \( v \) at that time. Calculating \( a \): \[ a = \frac{d^{2}s}{dt^{2}} = \frac{d}{dt}(24t^2 - 30t + 6) = 48t - 30 \] Now substituting \( t = 4 \): \[ a = 48(4) - 30 = 192 - 30 = 162 \, \text{m/s}^2 \] So the acceleration after 4 seconds is \( 162 \, \text{m/s}^2 \).