Answer
The particle is 32 m from the start point.
Work Step by Step
1. Using equation 12-1:
$$v = \frac{ds}{dt} \longrightarrow ds = v \space dt$$
$$\int_0^s ds = \int_0^tv \space dt$$
2. Substitute the values and integrate:
$$s = \int_0^t(4t - 3t^2)dt$$
$$s = \Big[2t^2 - t^3 \Big]_0^4 = (2(4)^2-(4)^3) - (2(0)^2 - (0))$$
$$s = -32 \space m \longrightarrow 32 \space m$$
