Explanation Given information: We have been given the following information: γ m = 1 .0, a = 32.2 ft s 2 , h = 32.0 inches, W = 3000 lb , L = 200 inch and μ = 1.0 . We have the following formula for the aerodynamic resistance, R a = ρ 2 C D A f V 2 Substituting the values of ρ = 0 .002378 slugs/ft 3 , C D = 0.35 and A f = 21 ft 2 , we have V = ( 10 mi h × 5280 ft 1 mi × 1 h 3600 sec ) V = ( 52800 ft 3600 sec ) V = 14.67 ft sec . R a = 0 .002045 slugs/ft 3 2 × 0.30 × 20 ft 2 × ( 14.67 ft sec ) 2 R a = 1.319 lb The value of coefficient of rolling resistance, f rl is as follows: f rl = 0.01 ( 1 + V 147 ) Put the value of V = 14.67 ft sec in the above equation, we have f rl = 0.01 ( 1 + 14 .67 147 ) f rl = 0.01 ( 147+14 .67 147 ) f rl = 0.01 ( 161 .67 147 ) f rl = 0.01 ( 1.0998 ) f rl = 0.011 For rolling resistance is given as follows: R rl = f rl W Put, f rl = 0.011 R rl = 0.011 × 3000 lb R rl = 32.99 lb R rl = 33.00 lb . For mass factor, we have the following formula: γ m = 1.04 + 0.0025 ε o 2 γ m = 1.04 + 0.0025 ( 4.5 ) 2 γ m = 1.04 + 0.0025 × 20.25 γ m = 1.04 + 0.0506 γ m = 1.0906 Now for engine-generative tractive force, we have the following formula : F e = M e ε o η d r Substitute the values, we have F e = 95 ft-lb × 4 .5 × 80 100 13 in 12 ft F e = 342 .00 ft-lb 1.0833 ft F e = 315.70 lb . For Front wheel drive case, we have F max = μ W ( l f + f r l h ) L 1 + μ h L Substitute the values, we have F max = 0

The acceleration for front and rear-wheel-drive.

Principles of Highway Engineering and Traffic Analysi (NEW!!)

6th Edition

ISBN: 9781119305026

Author: Fred L. Mannering, Scott S. Washburn

Publisher: WILEY

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