Question

Equations of Motion: Translation Learning Goal: To use the equations of motion as they relate to linear translation of an object to determine characteristics about its motion. The car shown has a mass of m= 1200 kg and a center of mass located at G. The coefficient of static friction between the wheels and the road is us 0.240. The dimensions are a = 1.05 m. b= 1.65 m, and c= -0.350 m Assume the car starts from rest, the wheels do not slip on the road, and that the car experiences constant acceleration. Neglect the mass of the wheels (Figure 1) Figure < 1 of 1 >

Answer 1

Here we are goint to use Motion equation to solve this problem.Please go through images given below for step by step answer of your problem,

Solution page-s F. B.D. or car W = 1200x 981 M 1200xar J c = 0.350m. 0.240NA 20240 NB NA 1.05 11.65 Given pata : m= 1200kg ls= 6.240 a- 1.05m bo 1.65m Co v. 350m. find forces and moments, Efx = 0.240NA+ 0.240MB T 1200 AG. + Efy = NA TNB - (1200x9.81) G+EMG 0 -Max 1,05) + COBX 165) -160,140NA +6.740 MB)(0:3)] 0.

Page 2 Rear 1 for fight when drive, 0.240 NB to put in egn. ② and 3 0.2400A = 1200dG. INAINB- (1200x9.81) = 0 1.2 -CNAX1.05) + 1.654B-[(0.240 NA) (0.35] - - 1.05NA & 16 SNB -0.084 NA=0 -1.134 NA + 1.65 NB=0. (13 solving above equations, NA-6976.9 70 HB=4795.) M 70 W=12oox 9.81 aq - 1. m 0.240 NA 0.240NB-o. S2 fig. b vi- mis V2 = 82 km/hr ? Given : V2 = 22.78 mls- Use equation of motion, Ne v taat. 22-78 = 0+14t. It= 16:2715) Ans. 6. which is is shortest time to reach speed of 82

Page 3 Front wheel drive, 0.2TONA-O. pur in echt. ③ and 3. 0.240 NB - 1200 la NA+NB- (1200x9.81)=0 - CNAX 105) + CNB X1.65) - ((0.240 NB) (0.35)] = o. - 1.05NA + 1.65NB - 0.084 MB=0 - 105NA +1.566 NB TO 3.2 Solving, NA=7047 NYO, NBE 4725M COK) dae 0.945 m S2 Vi omis Griven VE 82 km/hr. Given W = 1200x9 81 1200xOG aco 0.2460B converto mls. 0.24MATO V- 22.78 m fig, @ How use equoition of motion, 12 = , + dat. 22.78 = 0 +(6.445*2) = 24.105 5 7 Aps. -2 which is speed of shortest time it takes the car to reach Starting from rest. 82 km 1 hr

Page 4 ♡ for 4-wheel drive ( Please refer fig, @) 0-240 NA +0.240 NB = 12000 MA & HB - 1200X9.81 (23) 0.084 - 1.05NA + 1.65 NB - TONA -0.084 HBO - 1.134 NA +1.566NBO 3:3 HA - 6827.76 N 70 COK) NB - 49 44 244 aa= 2.35 m S2 How use equation of motion, V2=v&aat 278 -o +2.35 2- It = 9.6935 Ans. 3. which is shortest time to reach speed of 82 km br.