At time $t$, the position of a body moving along the $s$-axis is $s=-t^{3}+12 t^{2}-45 t m$.
a. Find the body's acceleration each time the velocity is zero.
b. Find the body's speed each time the acceleration is zero.
c. Find the total distance traveled by the body from $t=0$ to $t=4$.

Step 1 ：The position of the body is given by \(s=-t^{3}+12 t^{2}-45 t m\).

Step 2 ：The velocity of the body is the derivative of the position with respect to time, \(v = ds/dt\).

Step 3 ：Differentiating the position function, we get \(v = -3t^{2}+24t-45\).

Step 4 ：The acceleration of the body is the derivative of the velocity with respect to time, \(a = dv/dt\).

Step 5 ：Differentiating the velocity function, we get \(a = -6t+24\).

Step 6 ：a. To find the body's acceleration each time the velocity is zero, we set the velocity function equal to zero and solve for \(t\).

Step 7 ：Setting \(v = -3t^{2}+24t-45 = 0\), we can solve the quadratic equation to find \(t = 3, 5\).

Step 8 ：Substituting these values into the acceleration function, we find the acceleration when the velocity is zero: \(a(3) = -6*3+24 = 6, a(5) = -6*5+24 = -6\).

Step 9 ：So, the body's acceleration each time the velocity is zero is \(\boxed{6 m/s^{2}}\) and \(\boxed{-6 m/s^{2}}\).

Step 10 ：b. To find the body's speed each time the acceleration is zero, we set the acceleration function equal to zero and solve for \(t\).

Step 11 ：Setting \(a = -6t+24 = 0\), we find \(t = 4\).

Step 12 ：Substituting this value into the velocity function, we find the speed when the acceleration is zero: \(v(4) = -3*4^{2}+24*4-45 = 3\).

Step 13 ：So, the body's speed each time the acceleration is zero is \(\boxed{3 m/s}\).

Step 14 ：c. The total distance traveled by the body from \(t=0\) to \(t=4\) is the integral of the absolute value of the velocity function from 0 to 4.

Step 15 ：The velocity function \(v(t) = -3t^{2}+24t-45\) changes sign at \(t = 3\) and \(t = 5\), but we are only interested in the interval from 0 to 4.

Step 16 ：So, we split the integral into two parts: from 0 to 3 (where the velocity is positive) and from 3 to 4 (where the velocity is negative).

Step 17 ：The total distance is then \(\int_{0}^{3} (-3t^{2}+24t-45) dt + \int_{3}^{4} (3t^{2}-24t+45) dt\).

Step 18 ：Evaluating these integrals, we find the total distance traveled by the body from \(t=0\) to \(t=4\) is \(\boxed{18 m}\).