## Engineering Mechanics: Statics & Dynamics (14th Edition)

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
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: $$v_{avg} = \frac{8 \space m}{3 \space s} = 2.67 \space m/s$$