Answer
Particle's deceleration when t = 3 s: 12 $m/s^2$
Particle's position at the same instant: 0 m
How far has the particle traveled: 8 m
Average speed: 2.67 m/s
Work Step by Step
1. Using equation 12-2, find the acceleration:
$$a = \frac{dv}{dt} = \frac{d(6t - 3t^2)}{dt} = 6 - 6t$$
2. At t = 3 s:
$$a = 6 - 6(3) = -12 \space m/s^2$$
3. Deceleration is the negative of the acceleration:
deceleration = 12 $m/s^2$
4. Using equation 12-1, find the position of the particle:
$$v = \frac{ds}{dt} \longrightarrow ds = v \space dt$$
- Integrate:
$$s = \int_0^3(6t - 3t^2)dt = \Big[3t^2 - t^3 \Big]^3_0$$ $$s = 3(3)^2 - (3)^3 - (3(0)^2 - 0^3) = 0 \space m$$
5. Investigate the path of motion:
$v = 6t - 3t^2$.
Find the points where v = 0:
$$0 = 6t - 3t^2 \longrightarrow 0 = t(6 - 3t)$$
v =0 when t = 0 and when 6 - 3t = 0: $6 = 3t \longrightarrow t=2$
Therefore, we have to sum the amount traveled between 0 and 2 seconds, with the same between 2 and 3 seconds.
6. Integrate:
$$s_1 = \Big[3t^2 - t^3 \Big]^2_0 = 3(2)^2 - (2)^3 = 4 \space m$$
$$s_2 = \Big[3t^2 - t^3 \Big]^3_2 = 3(3)^2 - (3)^3 - (3(2)^2 - (2)^3)$$
$$s_2 = - 4 \space m$$
7. Find the total amount:
$$s_T = 4 + 4 = 8 \space m$$
8. Use the equation to find the average speed:
$$v_{avg} = \frac{8 \space m}{3 \space s} = 2.67 \space m/s$$