Physics question???

Question

A particle's velocity is described by the function V(t)=k t^2 m/s, where k is a constant and t is in s. The particle's position at t=0 is -9m and at t=3.0s, its position is +9m. Determine the value of constant k. Be sure to include the proper unit.

Answer 1

The particle's position is the integral of the velocity:
x = ∫(v(t))dt =∫(kt²)dt = (1/3)kt³ + C.
Your first constraint is that x(0) = -9m and x(3) = 9m. Plug this information in:
The first equation is
x(0) = -9 m = (1/3)k(0)³ + C or C = -9 m
The second equation is
x(3) = 9 m = (1/3)k(3)³ + C,
but C = -9 m
9m = (1/3)k(3)³ - 9m,
18 m = (27 sec³/3) k
2m/sec³ = k

Answer 2

If V(t) = k t^2 then since V(t) = dx/dt you need to integrate V(t).

x(t) = Integral(V(t) dt) = integral(k t^2 dt) = 1/3 kt^3 + constant

Now x(t=0) = -9 = constant

and

x(t=3) =9 =1/3 k (3)^3 - 9 = 9 k - 9

18 = 9 k ----> k =2 m/s^3

Answer 3

maybe integrate to find distance s(t)

s(t) = INT v(t) dt = 1/3 k t^3 + const

s(0) = 1/3 k 0^3 + const = -9 => const = -9 m

EDIT:
s(3) = 1/3 k 3^3 -9 = 9 => k = 18 * 3 / 27 = 2 ms^-3