Question #216077

A vehicle with a mass of 1 500 kg is accelerated with a constant acceleration from rest to a speed of 90 km/h up an incline of 1 in 20. The duration of acceleration is 25 seconds. The constant tractive resistance is 220 N. Calculate:

2.1. The engine power required if the efficiency is 80% at the instant when the vehicle reaches

90 km/h. (8)

2.2. The engine power required to move the vehicle at constant speed of 72 km/h the

efficiency is 82%. (6)


1
Expert's answer
2021-07-12T12:15:57-0400
0.8P=(220+1500(9.8)(0.05))(90)2(3.6)P=14.9 kW0.8P=\frac{(220+1500(9.8)(0.05))(90)}{2(3.6)}\\P=14.9\ k W

2)


0.82P=(220+1500(9.8)(0.05))(72)(3.6)P=23.3 kW0.82P=\frac{(220+1500(9.8)(0.05))(72)}{(3.6)}\\P=23.3\ k W


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