4 The position $s$ of a point (in miles) is given as a function of time $t$ (in minutes): $s=t^{3}-3 t^{2}+5 t$
a. Find its velocity as a function of time
b. Find its velocity at $t=2$ minutes
c. Find its acceleration as a function of time
d. Find its acceleration at $t=2$ minutes

Step 1 ：The position $s$ of a point (in miles) is given as a function of time $t$ (in minutes): $s=t^{3}-3 t^{2}+5 t$

Step 2 ：The velocity of the point is the derivative of the position function with respect to time. So, we need to find the first derivative of the given function. The velocity as a function of time is $v(t) = 3t^{2} - 6t + 5$

Step 3 ：We can substitute $t=2$ into the velocity function to find the velocity at $t=2$ minutes. The velocity at $t=2$ minutes is $v(2) = \boxed{5}$ miles per minute

Step 4 ：The acceleration is the derivative of the velocity function with respect to time. So, we need to find the second derivative of the given function. The acceleration as a function of time is $a(t) = 6t - 6$

Step 5 ：We can substitute $t=2$ into the acceleration function to find the acceleration at $t=2$ minutes. The acceleration at $t=2$ minutes is $a(2) = \boxed{6}$ miles per minute squared