3. A fixed point O lies on a straight line. A particle P moves along the straight line. At time t seconds, tgeqslant 0 the distance, s metres , of P from O is given by s=(1)/(3)t^3-(5)/(2)t^2+6t (a) Find the acceleration of P at each of the times when P is at instantaneous rest. (b) Find the total distance travelled by P in the interval 0leqslant tleqslant 4

# Explanation:## Step 1:Acceleration is the second derivative of the position function. To find the acceleration, we first derive the velocity function \(v(t)\) from the given position function \(s(t)\) by applying the derivative, then derive the acceleration function \(a(t)\) from this obtained velocity function. The position function is \(s(t) = \frac{1}{3}t^{3}-\frac{5}{2}t^{2}+6t\). Therefore, make two-time differentiation will give us acceleration function. ## Step 2:Let's differentiate the position function to get the velocity function:### \(v(t) = \frac{d}{dt}s(t) = \frac{d}{dt} ( \frac{1}{3} \cdot t^3 - \frac{5}{2} \cdot t^2 +6 \cdot t) = t^2 - 5t + 6\)## Step 3:Now we differentiate the velocity function to get the acceleration:### \(a(t) = \frac{d}{dt}v(t) = \frac{d}{dt} (t^{2} -5t + 6) = 2t - 5\)# Answer:The general acceleration function is:### \(a(t) = 2t - 5\)___# Explanation:## Step 4:Now for (a), we need to find the times when the velocity is equal to zero (i.e., instantaneous rest). These times can be found by setting \(v(t)\) to zero and solving for . Or equivalently, solving the quadratic equation .## Step 5:By solving the above equation, we will be able to have the times when at instantaneous rest.## Step 6:Then we substitute these values in the acceleration function to find the acceleration of the particle at the instant when the particle is at rest.# Answer:First, let's note that the roots/calculated-answers of the quadratic equation depend on its coefficients, related to the values of by logos equation associated with the discriminant adjusted from the quadratic equation. This result can be verified. We find the velocity is zero at two times and . The accelerations at these times appear in \(a(t_1)\) and \(a(t_2)\).____# Explanation:## Step 7:For the question part (b), the total distance traveled by in the interval is an integration question. The total distance is the absolute value of the integral of the velocity, since velocity may change the direction. For the value we calculated from step 3 (velocity function), we just Equate (Equalize) intervals when moving in one direction by calculating assorted velocity/time ranges, Then take their sum.## Step 8:Taking absolute we go part by part: We first focus at [for intervals negatively going points [- , )]("negative direction")* and [We second focus at intervals positively going ("positive direction")**], Deduce all positive roots associated with negative going and negative roots & zero associated with positive going.## Step 9:Perform a mathematical operation on the final absolute integration result and display both the processed and unprocessed results on the processed result # Answer:From the jump inside the integral (1st rectify), Conservation on the jumps need to clear isolated roots under absolute integral. The answer should always be labelled with positive valuing.Ultimately, taking the whole value sum for all, Ultimately, **********a longer route expected since more procedure goes through: [Total distance = Integral on abs( .] # Student Simplification...integrated on 0-4 segments couch up those varied as∫|₀⁴ $|dx ..., causes the equation with extra calculations has no equal trim-regionality, drawn between wholes and fractions serving sufficient time adjusting calculated uptime beginners learn during an overwhelm.