A vehicle with a total weight of 3000 lbf has a 50-50 weight distribution, front
ID: 2325200 • Letter: A
Question
A vehicle with a total weight of 3000 lbf has a 50-50 weight distribution, front to rear. The wheelbase is 100 inches and vertical center of gravity is 31 inches. The vehicle accelerates from zero to its maximum speed in a set distance of 1320 feet. The vehicle is a front wheel drive and has an independent front suspension and drive axle. The peak friction coefficient between the tire and the pavment is 0.8. If the vehicle operates in such a way as to allow acceleration at its traction limits, calculate the maximum possible vehicle velocity, in mph, in this distance.
Explanation / Answer
In static condition the weight is shared by front and rear tyres. Hence the front tyres take 1500 lbf.= 6720 N
As the vehicle accelerates the rear wheels are loaded = ( 31/100) x M Xacceleration where mass is 13440N
= 4166.4 x a
hence the front wheel load is 6720 - ( 4166..4 xa )
This should be equal to the traction of the tyre ie. 672 0 x o.8 = 5376 N
Therefore a = 6720- 5376 / 4166.4 = 0.3225 m/sec2
S = ut + 0.5 at2 u = 0
hence we have s= 1320 feet = 402.3 meter = 0.5 x o.3225x t2
t = sqrt ( 402.3 x 2/ o.3224 ) = 49.95 sec.
V = at == 0.3225 x 49.95 = 16.10 M/sec = 36 MPH