How to justify the anwers? problem # 2.10 (mannering 43)resistance, tractive effort, and acceleration (sections 2.2-2.7)an engineer designs a rear-wheel–drive car (without an engine) that weighs 2000 lb and has a 100- inch wheelbase, drivetrain efficiency of 80%, 14-inch– radius wheels, an overall gear reduction ratio of 10 to 1, and a center of gravity (without engine) that is 22 inches above the roadway surface and 55 inches behind the front axle. an engine that weighs 3 lb for each ft-lb of developed torque is to be placed in the front portion of the car. calculations show that for every 20 lb of engine weight added, the car’s center of gravity moves1 inch closer to the front axle (but stays at the same height above the roadway surface). if the car is starting from rest on a level paved roadway with a coefficient of adhesion of 0.8, select an engine size (weight and associated torque) that will result in the highest possible available tractive effort. problem 2-10hint: for highest tractive effort, max. tractive should equal engine generated tractive effort. get, fe (equ. 2.17) fmax. (equ. 2.14) weng = 3me 3 ibs per 1 unit of torquelfnew = lf –weng/20 1 inch per 20 ibs. of engine weight. wtotal = w0 + weng. setting fe = fmax. solve for me. answers. me = 122.152 ib. weng = 366.456 ib. wtotal = 2366.50 ib. book used for the problem is principles of highway engineering and traffic analysis (5th edition)