Answer
The acceleration when t = 4s is equal to -24 $m/s^2$, the displacement from t =0 to t = 10 s is equal to -880 m, and the distance the particle travels during this time period is 912 m.
Work Step by Step
1. Using equation 12-2:
$$a = \frac{dv}{dt} = \frac{d(12 - 3t^2)}{dt} = -6t$$
2. Substitute t = 4s:
$$a = -6(4) = -24 \space m/s^2$$
3. Using equation 12-1:
$$v = \frac{ds}{dt} \longrightarrow v \space dt = ds$$
4. Integrate:
$$\int_0^{10}(12 - 3t^2)dt = \int_0^s ds$$
$$\Big[12t - t^3\Big]^{10}_0 = s$$ $$12(10)- (10)^3 = s$$ $$s = -880 \space m$$
5. Investigate the path of motion:
$$v = 12 - 3t^2$$
At t = 0, v = 12. (Positive).
$v = 0$ at:
$$0 = 12 - 3t^2 \longrightarrow 12 = 3t^2 $$ $$4 = t^2$$ $$t = 2\space s$$
Therefore, $v$ is positive when $0 \lt t \lt 2$, and positive when $2 \lt t \lt 10$
6. Find the total distance traveled by the particle:
$$ s_1 = \Big[12t - t^3\Big]^{2}_0 = 12(2) - (2)^3 = 16 \space m$$ $$s_2 =\Big[12t - t^3\Big]^{10}_2 = 12(10) - (10)^3 - (12(2) - (2)^3) $$ $$s_2 = -896 \space m$$ $$s_T = 16 + 896 = 912 \space m$$
