(a) if a particle’s position is given by x 4 12t 3t 2 (where t is in seconds and x is in meters), what is its velocity at s? (b) is it moving in the positive or negative direction of x just then? (c) what is its speed just then? (d) is the speed increasing or decreasing just then? (try answering the next two questions without further calculation.) (e) is there ever an instant when the velocity is zero? if so, give the time t; if not, answer no. (f) is there a time after t 3 s when the particle is moving in the negative direction of x? if so, give the time t; if not, answer no.

A.) I think the equation is written as: x = 4(12t + 3t²). Since there are missing operations, I think that is the logical form of the equation. Let's simplify it by distribution.

x = 48t + 12t²

The velocity or speed is the derivative of x with respect to t. 
dx/dt = v = 48 + 12(2)t = 48 + 24t
Thus, the velocity at any time t is given by the equation: v = 48 + 24t.

B.) This depends at what time. But because the operation only involves addition, it is impossible to yield a negative answer. So, the answer would be positive indicating that it travels in the +x direction.

C.) The speed is just equal to velocity, apart from the positive or negative sign. Thus, if at time t = 1 s, the velocity is equal to +72 m/s. Then, the speed is also 72 m/s.

D.) As time increases, the speed will also increase because of the addition operation. If it were subtraction, then the speed will decrease.

E.) Let's see if we equate v = 0.
48 + 24t = 0
t = -2

Since it is impossible to get a negative value of time, then it is impossible to have a velocity of zero.

F.) Replace t=3 to the equation:

48 + 24(3) = 120 m/s
Because the answer is positive, then it is impossible to travel in the negative x direction.