Vehicleperformance assignment 2
Arear wheel drive car of mass 2200kg has a rolling resistance of 150N, a frontal area of 1.9m2, a drag coefficient of 1.00, atransmission loss of 7% and a maximum engine power of 59kW. Assumingthat the traction limit for each tyre is equal to the weight that itcarries (i.e. the tyre traction coefficient =1) and that 60% of thevehicle weight is on the rear wheels during acceleration. Take thedensity of air to be 1.2 Kg/m3 and g = 9.8 m/s2Calculations
Mass=2200kg
RollingResistance=150N
Frontalarea=1.9 m^{2}
DragCoefficient =1.00
Transmissionloss=7%
MaximumEngine power=59kW
Tyretraction coefficient =1
Weighton the rear wheel =60%
Densityof air 1.2kg/m^{3}
g=9.8 m/s^{2}

What is the maximum power available at the back wheels?
Drivetrain efficiency Enginepower = (10.07)59 kW=54.87kW=54870 Watts

What is its theoretical maximum speed in mph on a level road?
Rearwheels power= Aerodynamic power loss + Rolling resistance + powerinto hills
Rollingresistance power = Rolling resistance Force velocity = 150 N V
AerodynamicPower loss =Drag Velocity=
Poweron level ground = 0
54,870Nm/s = 150V + ()
54,870= 150 V + 1.14V^{3}
Toquickly solve for V I will use iteration method:
35.1m/s…. 54,562.65 Nm/s
35.15m/s…. 54,781.12 Nm/s
35.16m/s …54,824 Nm/s
35.17m/s ….54,868.67 Nm/s
35.18m/s ….54,912.49 Nm/s
So78.67mph (35.17 m/s) is the fairly accurate top speed.

What is its theoretical maximum speed in mph up a 1 in 4 hill?
54,870Nm/s= Rolling resistance + aerodynamic power loss + power into hills
Powerinto hills (1 in 8 hill) = mass g change in height per unit time
Powerinto hills (1 in 8 hill) =2200 9.8 (1/4) = 5,390
Substitutingin the above equation
54,870Nm/s = 150V + 1.14 + 5,390
Toquickly solve for V I will use iteration method:
33.6m/s …53,673.68 Nm/s
33.7m/s …54,075.93 Nm/s
33.8m/s …54,480.49 Nm/s
33.9m/s …54,887.37 Nm/s
So75.83 mph (33.9 m/s) is the approximate top speed up 1 in 8 hill.

What is the maximum tractive force that each of the rear tyres could provide under acceleration before slipping occurs? (NB this is per rear wheel not per axle!!)
Tractiveforce = Coefficient of friction Downward force
Tractionforce = 1(0.6022009.8) = 12,936N

What would be the maximum, useable, tractive force at each of the rear tyres if full power was applied at 20 mph? (Hint: is this limited by tyre grip or by engine power?)
Useabletractive force will be the minimum of (force due to engine power,tire traction force)answer = min(54870Nm/s / 8.94m/s , 12,936N)answer = min(6,137.58 N, 12,936 N) = 6,137.58N

What would be the wind resistance at 20 mph?
Windresistance =
=91.11N

What would be the acceleration (in mph/s) at 20 mph on the level?
Force(F) = Tractive force – Aerodynamic Drag Rolling resistance force
AerodynamicDrag = 91.11N
Tractiveforce =6,137.58 N
Rollingresistance force =150 N
6,137.58N – 91.11 N – 150N = 5,896.47 N
=5.995MPH/s
Whatwould be the acceleration (in mph/s) at 20 mph up a 1 in 4 hill?
Force(F) = Tractive force – Aerodynamic Drag Rolling resistanceGravity
Where
AerodynamicDrag = 91.11N
Tractiveforce =6,137.58 N
Rollingresistance force =150 N
Gravitationpull due to the hill = mass g (1/4) = 22009.8(1/4)= 5390N
6,137.58N – 91.11 N – 150N 5,390 N= 506.47 N
=0.2302=0.515MPH/s