Question

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?

Answer 1

Answer:

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

Answer 2

Please see the picture below for the methodology and results.