Question

6. A train 300m long passes through a tunnel 210m long. The interval between the entrance ocity of of the locomotive and the exit of the last wagon is 17s. Calculate the average vel the train 7. Complete the following table: Moving object Average speed Average Distance Time interval speed (Km/h) Snail Runner Car irplane aglev 16mm 100m 10s 30 min 2h 30min 50km/h 1800km 200km 400km/h

Answer 1

Length of train L_T_r = 300 m Length of tunnel L_T_r = 210 m time interval to go through the tunnel t = 17 s Therefore total distance covered by the train in the t, & L = L_T_r + LT_n = 300 + 210 = 510 m Therefore avg velocity of the train v = 1/t = 510/17 m/s = 30 m/s Therefore V = 30 m/s \r\n avg speed = distance/time interval snail, avg speed = 16 mm/9s = 16 times 10^-3 m/9 s = 4 times 10^-3 m/s = 4 times 10^-3 times 3600/1000 Km/h = 1.44 times 10^-2 Km/h Runner, avg speed = 100 m/10 s = 10 m/s = 36 Km/h car, abg speed = 50 km/h = 50 times 1000/3600 times m/s = 13.89 m/s \r\n distance covered in 30 min or 1/2 hour = 50 times 1/2 km = 25 Km Airplane, avg speed = 1800 km/2.5 km = 720 km/h = 200 m/s