Question

Physics topics: Newton's Law (F m a); definitions of a, V & x graph F, a, V, x, t Statics topics: Equilibrium (EF-0 & EM-0); distributed loads; free-bodies; vector components Thermo topics: Mass & energy balances; incompressible liquids; ideal gases; isentropic process 1a.) A 1500 kg automobile accelerates from "zero to sixty" miles per hour (0 to 27 m/s) in 8 seconds. Assume constant acceleration and no friction. Graph the power required (P), the speed (V) and the distance traveled (x) versus time (t). Find numeric values att-8 seconds. At time, t 8 seconds, F P (kw) N,P= kW,V = 27 m/s, x= 一 V (m/s) x (m) 8 4 What do the slopes represent? What do the areas under the curve represent? 1b.) Repeat assuming constant power in place of constant acceleration. (P AKE/At 68.3 kW) At time, t = 8 seconds, F = N, a = m/s, V = 27 m/s, x = m. F (N) V (m/s) x (m) HINT Newton's laws of Mechanics Work # ΔΚΕ a dVidt, integrate a to get V V- dx/dt, integrate V to get x t (s) I t (s) 4 4 4 0 What do the slopes represent? What do the areas under the curve represent?

Answer 1

V = u + at implies a = v - u/t = 27 - 0/8 = 3.375 m/s^2 p = FV = ma(u + at) F/t = 8 = 1500 times 27/8 = 5062.5N p/t = 8 = 1500 times 27/8 times 27 = 136.68 KW x = ut + 1/.2 at^2 = 1/2 times 27/8 times 8^2 = 108 M slope = ma^2 or Fa area = total power required area = distance travelled by car x = ut + 1/2 at^2 slope = velocity p = F = mav = 68.3 KN m v dv/dt = 68.3 KN m integral v dv = 68.3 KW integral dt mv^2/2 integral^v_0 = 68.3 [t]^t_0 v^2/2 times 1500 = 68.3 times 10^3 t implies v^2 = 91 t v = 9.5 squareroot implies dx/dt = 9.5 squareroot integral^x_0 dx = 9.5 intgeral^t_0 squareroot dt x = 9.5/3/2[t^3/2]^t_0 = 6.33 t^3/2 F = ma = mdv/dt = m times 9.5 times 1/2 squareroot t = 7125/squareroot N V/8 = t = 26.87 m/s x/t = 8 = 143.23 m & a/t = 8 = 1.68 m/s^2 = 27 m/s F/t = 8 2519 N