Question

Learning Goal: Part A - Shortest Time to Reach a Given Speed with Rear-Wheel Drive 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 = 1100 kg and a center of mass located at G. The coefficient of static friction between the wheels and the road is pls = 0.230. The dimensions are a = 1.25 m, b= 1.55 m, and c= 0.290 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. Determine the shortest time it takes the car to reach a speed of v = 82.0 km/h, starting from rest, if the engine drives only the rear wheels. Express your answer to three significant figures and include the appropriate units. View Available Hint(s) (Figure 1) Templates Symbols uñdo redo fedetl keyboard shortcuts help t= Value Units Submit Part B - Shortest Time to Reach a Given Speed with Front-Wheel Drive Determine the shortest time it takes the car to reach a speed of v = 82.0 km/h, starting from rest, if the engine drives only the front wheels. Express your answer to three significant figures and include the appropriate units. ► View Available Hint(s) Templates Symbols undo ređe redet keyboard shortcuts help t= Value Units Submit

Answer 1

Answer (a) with Rear-wheel drive FBD ag 0.29m Q ㅈ ta NB NA 1 1.55 m 1.25m frictional force = fa = 0.23 NA SNA - O →E Fz= m@ase FA = maga 0.23 NA =1100@gmx NA = 4782.608 agz → EFy=mary * NA UNB – 1100 (9.81) → 4782 .608 @Ge + No-1100(9.81) = 0 NB= 10791 - 4782.608G. →EMA = ({ MAJA NB(1-25+1.55)– W(1.25) = -mag e(0.29) from eq ④ [10791 - 4782.608@y](2.8)-(100(9.81)) (1.25) =-1100 (AG)x (0.29) → @aze = 1.279 m/s

we know, v=utat 82 C) = 0 + (1-279)t ( 1 Kimya ms ) t = 17.809 sec (b) with front-wheel drive OG G O 0.29 m A SfB NA NB - K 1.25m 1.55m -> fB = Ms NB = 0.23 NB Efx =mlaghe FB = maga 0.23N3 = 1100@6x => NB = 4782.608 @)& Efy=maro NA + NB -(1100X9.81) = 0 » NA TNB = 10791 NA + 4782.608@gre = 10791 NA = 10791 – 4782.608 Code B

EMB= ({MK)B - NAC1-25+:55) + (1.55) =-m@g)x(0.29) from eq -(10791– 4782.608QG 2)(2-3) + (1100x9.81X1.55) - 100@ax)0.29) (as) a= 0.983 m/s こー v=cet at 82% ( ) =0 + (0.983)t t = 23.171 sec. Ars (c) will All-wheel drive. (90) تبع T 0.29m Q Q T fo A NB 1.25m 1.55 m E Fx = m nao De Fatfo= max => 0.23 NA +0.23 NB = 1100@6)x NA +NB = 4782.608@gmx =>

e fy. = mcanly NA+ NB -(100X9.81) = 0 from eq 4782.668@x-(1100x9-81) = 0 @Qx= 2.256 m/s? v=etat 826) =0+2.256t t=10.096 sec aus Good Luck Please like.