Answer:
The number of "gs" of acceleration the vehicle must experience in order for the front wheels to lift off the ground is 1.844 gs.
Step-by-step explanation:
To solve the question, we assume that the center of gravity is acting at the center midway of the car, half the wheel base. That is at 93/2 inches
Since the weight is acting at the center of gravity then we have taking moment about the rear wheel gives
(5000 - 2542) × 7.75 + 5000 × 7.75/2 = 5000/32.2 × a × 4.166667
Which gives
38424.5 = 646.99798×a
a = 38424.5 /646.99798 = 59.39 f/s² which is equivalent to
59.39 /32.2 or 1.844 gs