1. **Problem statement:** A train starts from rest with constant acceleration. It travels at 30 ft/s at some point, and 160 ft farther it travels at 50 ft/s. We need to find: (a) acceleration, (b) time to travel the 160 ft, (c) time to reach 30 ft/s, (d) distance traveled before reaching 30 ft/s. 2. **Relevant formulas:** - Velocity under constant acceleration: $$v = u + at$$ where $u$ is initial velocity, $v$ is final velocity, $a$ is acceleration, $t$ is time. - Distance under constant acceleration: $$s = ut + \frac{1}{2}at^2$$ - Another useful formula: $$v^2 = u^2 + 2as$$ 3. **Given:** - Initial velocity $u = 0$ (train starts from rest) - Velocity at point 1: $v_1 = 30$ ft/s - Velocity at point 2: $v_2 = 50$ ft/s - Distance between points 1 and 2: $s = 160$ ft --- **(a) Calculate acceleration $a$:** Using $$v^2 = u^2 + 2as$$ between points 1 and 2, initial velocity is $v_1=30$ ft/s, final velocity $v_2=50$ ft/s, distance $s=160$ ft: $$50^2 = 30^2 + 2a(160)$$ $$2500 = 900 + 320a$$ $$2500 - 900 = 320a$$ $$1600 = 320a$$ $$a = \frac{1600}{320} = 5$$ ft/s² --- **(b) Time to travel 160 ft between 30 ft/s and 50 ft/s:** Using $$v = u + at$$ with $u=30$, $v=50$, $a=5$: $$50 = 30 + 5t$$ $$50 - 30 = 5t$$ $$20 = 5t$$ $$t = \frac{20}{5} = 4$$ seconds --- **(c) Time to reach 30 ft/s from rest:** Using $$v = u + at$$ with $u=0$, $v=30$, $a=5$: $$30 = 0 + 5t$$ $$t = \frac{30}{5} = 6$$ seconds --- **(d) Distance traveled before reaching 30 ft/s:** Using $$v^2 = u^2 + 2as$$ with $u=0$, $v=30$, $a=5$: $$30^2 = 0 + 2 \times 5 \times s$$ $$900 = 10s$$ $$s = \frac{900}{10} = 90$$ ft --- **Final answers:** - (a) Acceleration $a = 5$ ft/s² - (b) Time to travel 160 ft = 4 seconds - (c) Time to reach 30 ft/s = 6 seconds - (d) Distance before 30 ft/s = 90 ft