2.93: The velocities are 2 3 2 and 2 t t v t v B A δ γ β α-= + = a) Since v B is zero at t = 0, car A takes the early lead. b) The cars are both at the origin at t = 0. The non-trivial solution is found by setting x A = x B , cancelling the common factor of t , and solving the quadratic for [ ] . 4 ) ( ) ( 2 1 2 αδ--±-= t Substitution of numerical values gives 2.27 s, 5.73 s. The use of the term “starting point” can be taken to mean that negative times are to be neglected.
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