A vehicle with a total weight of 3000 lbf has a 5050 weight

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.

Solution

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