The velocity of a particle traveling in a straight line is given by v = (6t - 3t^2) m/s, where t is in seconds. Assume s = 0 when t = 0. Part A: Determine the particle's deceleration at t = 3.1 s. Part B: Determine the particle's position at t = 3.1 s. Part C: How far has the particle traveled during the 3.1-second time interval? Part D: What is the average speed of the particle for the 3.1-second time interval?

Part A: To determine the particle's deceleration at t = 3.1 s, we need to calculate the acceleration at that point in time. Acceleration a(t) is the derivative of velocity v(t) with respect to time t. Given the velocity of the particle as v(t) = (6t - 3t^2) m/s, we will differentiate it to get the acceleration.

a(t) = dv(t)/dt = d/dt (6t - 3t^2) = 6 - 6t

Now we can plug in t = 3.1 s:

a(3.1) = 6 - 6(3.1) = 6 - 18.6 = -12.6 m/s^2

The negative sign indicates that the particle is decelerating. The deceleration at t = 3.1 s is 12.6 m/s^2.

Part B: The position of the particle is obtained by integrating the velocity function, since velocity is the derivative of position with respect to time. The integral of v(t) will be the position function s(t), and since s = 0 when t = 0, we know the constant of integration will be 0.

s(t) = ∫v(t) dt = ∫(6t - 3t^2) dt = (3t^2 - t^3) + C

With the initial conditions s(0) = 0, C = 0 and our position function simplifies to:

s(t) = 3t^2 - t^3

Now we can determine the position at t = 3.1 s:

s(3.1) = 3(3.1)^2 - (3.1)^3 = 3(9.61) - 29.791 = 28.83 - 29.791 ≈ -0.961 m

So the particle’s position at t = 3.1 s is approximately -0.961 m, indicating that it has passed through the origin (starting point) and moved in the opposite direction.

Part C: To find out how far the particle has traveled during the 3.1-second time interval, we need to find out when the particle changes direction (when the velocity becomes zero), because the position function indicates the particle has moved back towards the starting point.

The velocity equals zero when:

0 = (6t - 3t^2) = 3t(2 - t)

t = 0 (we already know this) or t = 2 seconds.

This tells us that the particle changes direction at t = 2 seconds. The position at t = 2 seconds is:

s(2) = 3(2)^2 - (2)^3 = 12 - 8 = 4 meters.

The particle has moved 4 meters from the start to its turning point, and then it starts moving back. At t = 3.1 seconds, we already found the position to be approximately -0.961 m, which means it is 0.961 meters from the starting point in the opposite direction.

Thus, the total distance traveled is the distance traveled until the turnaround point, plus the distance it returns:

Distance traveled = 4 m + 0.961 m ≈ 4.961 m.

Part D: The average speed of the particle for the 3.1-second time interval is the total distance traveled divided by the time interval.

Average speed = Total distance / Time interval

Average speed ≈ 4.961 m / 3.1 s ≈ 1.6003 m/s.

So, the average speed of the particle during the 3.1-second time interval is approximately 1.6003 m/s.