For this type of problem you need to find the first and second derivative of position, which represent velocity and acceleration respectively.
𝑠'(𝑡) = 3𝑡2 − 18𝑡 <<< this is the velocity of the particle as a function of time, t.
𝑠"(𝑡) = 6𝑡 − 18 <<< this is the acceleration of the particle as a function of time, t.
Question 1: Velocity at t = 4, this is asking you to evaluate s'(t) at t = 4. Here goes:
s'(t = 4) = 3(4)2 - 18(4)
= 48 - 72 = -24 ft/s
Question 2: Acceleration at t = 4, this is asking you to evaluate s"(t) at t = 4. Here goes:
s"(t = 4) = 6(4) - 18
= 24 - 18 = 6 ft/s2
Question 3: Since at t = 4, the velocity and acceleration have opposite signs, I conclude that the particle's speed is decreasing.
Question 4: To find when the particle changes direction set s'(t) = 0 and find any time when t > 0. Here goes:
s'(t) = 0 = 3𝑡2 − 18𝑡
0 = 3t(t - 6)
t = 0 and t = 6
The two solutions are t = 0 and t = 6, since we are looking for a time t > 0 in which the particle changes direction t = 6 s is when the particle changes direction.
If the particle was moving on the x axis and a negative velocity means it was going left on the x-axis. After 6 s it will change to going right.