A vehicle with a total weight of 3000 lbf has a 50-50 weight distribution, front

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

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