The 1997 Dodge Viper has mass 1547 kg and an engine that can produce a maximum driving force of 12.36 kN. Suppose drag is proportional to speed such that k = 164 Ns/m. (a) Determine the initial acceleration of the Viper from rest. (b) Determine the maximum speed (i.e. terminal velocity). (c) Calculate the distance and speed for t = 12.2 s. (d) Find time and distance for the car to reach 99% of its top speed. So basically I found the answers for a and b they are as follows. A)F=ma 12360N=1547kg(a) a=7.99N/kg B)F=ma Fcar-Fdrag= 0 Fcar=Fdrag F=kv 12360N=164(v) v=75.37 So those two are checked and fine but c im havin trouble still c) F-Fd=ma F-kv=ma F-kv=m dv/dt integral 1/m dt= integral dv/(f-k(v)) t/m= integral dv/(f-kv) so im stuck here. o and i think the integral limit is 0 and V. my teacher said to do u substitution and make u=F-kv and du=-kdv thereby making t/m= -1/k integral du/u with limit f and f-kv i have no idea what to do next with the integration. Basically i need someone to help integrate it and solve for v. then with that equation i can solve for part d. also i have no idea how to calculate the distance. o i need this for tomorow morning so any quick help would be very helpful thanks.