Mechanical systems - Please help me with this question! :( Person_Number M (kg) D (m) R (m/s2) <431383> 2700 60 1.4 Mechanical Systems Engineering Assignment 1 An unloaded lorry of mass M kg accelerates from rest and reaches a velocity of 30mph in 10 seconds. It then travels at constant velocity for half a minute before the driver applies the brakes and brings the vehicle to rest over a distance of D m. The lorry is then loaded with 800 kg of building materials that need to be transported over a total distance of 0.6 km. After loading, the lorry accelerates over a distance of 80m to attain a velocity of 30mph again. The vehicle subsequently maintains constant velocity before decelerating to rest at a rate of R m/s2. Using only your own personal input data as given on the assignment 1 “input values” spreadsheet (penalties will result from not using your OWN data):- a) Find for the unloaded lorry: ( 6 marks) i) the force required to accelerate the vehicle, and the retarding force necessary to bring it to rest; ii) the total time and the total distance travelled during this period. b) Find for the loaded lorry : ( 7 marks) i) the force required to accelerate the vehicle and the retarding force necessary to bring it to rest; ii) the total time taken during this period and the distance travelled at constant velocity. c) Calculate the % change in the momentum of the lorry from loaded to unloaded, whilst travelling at the specified constant velocities. ( 3 marks) d) Draw the ‘velocity-time’ graph of the motion of the lorry in its unloaded and loaded states, indicating clearly salient points and values. ( 4 marks) Relevant equations: S=(u+v/2)t S = Displacement V = velocity A= acceleration F= ma Any help on this question would be extremly grateful as its just a little assignment i need to do - i dont have a clue how too :S I also have to sketch a VxT graph but i will be able to handle that haha :P any help would be grateful cheers guys Regards mark The Work i have done so far is a = (v-u/t) = 30 - 0 /10 = 3ms^2 F = mxa = 2700 x 3 = 8100 N 8100 is the force required to accelerate? Retarding force : m(-a) = 2700 x (-3) = -8100N ?? Total time.. = 50 seconds? distance traveld 15x50/2 = 375m ?? Am i correct or have i got it totally wrong???