Engineering Physics (PHY1143) Assignment 3 Solutions 1. Xi = 0 Xf = 11” = 0.28 m Vi = 250 miles/hour = 111.76 m/s Vf = 0 (assuming the bullet stops) a = t = (Vi + Vf) / 2 = (Xf – Xi) / t (111.76 + 0) / 2 = (0.28 – 0) / t t = 0.0048 s a = (Vf – Vi) / t a = (0 – 111.76) / 0.0048 a = ‐23470 m/s² a = ‐23000 m/s² (considering significant figures)
2. Xi = 0 Xf = 20 * 3.50 m = 70.0 m Vi = 0 Vf = a = t = 25.0 s (Vi + Vf) / 2 = (Xf – Xi) / t (0 + Vf) / 2 = (70.0 – 0) / 25.0 Vf = 5.60 m/s a = (Vf – Vi) / t a = (5.60 – 0) / 25.0 a = 0.224 m/s² 3. The three things that the two cars share in this problem are the starting point, the point where they are together again (Xf), and the time (t). Car 1 (the speeder): Xi = 0 Xf = V1i = 160 kph = 44.4 m/s V1f = 160 kph = 44.4 m/s a1 = 0 (there is no change in velocity, therefore there is no acceleration). t = (V1i + V1f) / 2 = (Xf – Xi) / t (44.4 + 44.4) / 2 = (Xf – 0) / t Xf = 44.4 t … 1 Car 2 (the police car): Xi = 0 Xf = V2i = 0 V2f = a2 = 4.6 m/s² t = (Vi + V2f) / 2 = (Xf – Xi) / t (0 + V2f) / 2 = (Xf – 0) / t V2f = 2 Xf/t …. 2 a = (V2f – Vi) / t 4.6 = (V2f – 0) / t V2f = 4.6 t … 3 Substitute 1 into 2 V2f = 2(44.4 t) / t V2f = 88.8 m/s Substitute into 3 88.8 = 4.6 t t = 19.3 s Substitue back into 1 Xf = 44.4 (19.3) Xf = 856.2 m Xf = 860 m (that is 320 kph … is this reasonable?) 4. Xi = 0 Xf = 320 m Vi = 0 Vf = a = 9.81 (positive because we said down was positive) t = (Vi + Vf) / 2 = (Xf – Xi) / t (0 + Vf) / 2 = (320 – 0) / t Vf = 640 / t … 1 a = (Vf – Vi) / t 9.81 = (Vf – 0) / t Vf = 9.81 t … 2 Solve 1 = 2 640 / t = 9.81 t t² = 65.24 t = 8.1 s Substitute (either 1 or 2) Vf = 9.81(8.1) = 79.2 m/s Vf = 79 m/s (or 290 kph) with 2 sig. figures 5. Xi = 0 Xf = Vi = 53 m/s Vf = a = ‐9.81 m/s² t = 3 min. = 180 s (Vi + Vf) / 2 = (Xf – Xi) / t (53 + Vf) / 2 = (Xf – 0) / 180 Vf = Xf / 90 – 53 … 1 a = (Vf – Vi) / t ‐9.81 = (Vf – 53) / 180 Vf = ‐ 1718.8 m/s (negative because up is positive) Substitute into 1: ‐1718.8 = (Xf/90) – 53 Xf = ‐ 149382 m (or the rocket started 150 km above the ground) Xf = ‐ 150 km 6. The two things that the two cars share in this problem are the point where they get together (yikes) (Xf), and the time (t). Cheetah: Xi = 0 Xf = V1i = 0 m/s V1f = a1 = 5.4 m/s² t = (V1i + V1f) / 2 = (Xf – Xi) / t (0 + V1f) / 2 = (Xf – 0) / t V1f = 2 Xf / t … 1 a = (V2f – Vi) / t 5.4 = (V1f – 0) / t V1f = 5.4 t … 2 Solution 1 = 2 2 Xf / t = 5.4 t Xf = 2.7 t² … 3 Wildebeest: Xi = 120 Xf = V2i = 0 V2f = a2 = 2.1 m/s² t = (Vi + V2f) / 2 = (Xf – Xi) / t (0 + V2f) / 2 = (Xf – 120) / t V2f = 2 Xf/t – 240/t …. 4 a = (V2f – Vi) / t 2.1 = (V2f – 0) / t V2f = 2.1 t … 5 Solution 1 = 2 2 Xf / t – 240/t = 2.1 t Xf = 1.05 t² + 120 … 6 Substitute 3 = 6 2.7 t² = 1.05 t² + 120 1.65 t² = 120 t = 8.5 s Substitute (into equation 3 or 6): Xf = 2.7 (8.5)² Xf = 196 m Xf = 200 m from where the cheetah started the chase