determine the magnitude and units of the constant k. What is the net displacement of the particle over the same 6-second interval of motion?

54 m

Explanation:

Compute v (t):

a = dv / dt = - k*t^2

dv = - k*t^2. dt

Integrating both sides:

v = - kt^3 / 3 + C

v (0) = 12

Hence,

v (t) = 12 - k*t^3 / 3

Compute k:

@ t = 6s v (t) = 0

12 - k*t^3 / 3 = v (t)

12 - k * (6) ^3 / 3 = 0

k = 12 / 72 = 1 / 6 m/s^2

Compute s (t)

v = ds / dt = 12 - t^3 / 18

ds = (12 - t^3 / 18). dt

Integrating both sides

s (t) = s (0) + (12*t - t^4 / 72)

Net displacement:

s (6) - s (0) = 12*t - t^4 / 72

= 12*6 - 6^4 / 72

= 54 m