Engineering Mechanics: Statics & Dynamics (14th Edition)

1. Using equation 12-1: $$v = \frac{ds}{dt} = \frac{d(2t^2 - 8t + 6)}{dt} = 4t - 8$$ 2. Substitute v = 0, and solve for t: $$0 = 4t - 8 \longrightarrow 4t = 8$$ $$t = 2$$ 3. Investigate the path of motion. At t = 0: $$v = 4(0) - 8 = - 8 \space m/s$$. The velocity is negative. We know that v = 0 when t = 2, therefore: The velocity is negative between 0 and 2 second, and positive between 2 and 3 seconds. 4. Using equation 12-2, find the acceleration: $$a = \frac{dv}{dt} = \frac{d(4t - 8)}{dt} = 4$$ 5. Using equation 12-1 again: $$v \space dt = ds$$ $$s = \int_{t_0}^t v \space dt$$ $$s = \int^t_{t_0} (4t - 8) \space dt = \Big[2t^2 - 8t\Big]^t_{t_0}$$ 6. Calculate the distance traveled when v is negative $(s_1)$, and when it is positive $(s_2)$ $$s_1 = \Big[2t^2 - 8t\Big]^2_{0} = 2(2)^2 - 8(2) = -8 \space m$$ $$s_2 = \Big[2t^2 - 8t\Big]^3_{2} = 2(3)^2 - 8(3) -(2(2)^2 - 8(2)) = 2 \space m$$ 7. Add both distances to find the total distance traveled: $$s_T = 8 + 2 = 10 \space m$$