The distance of a car from a certain intersection after t seconds is given by d(t) = 12 + 3t - t2. When (that is, for which values of t) is the car 12 units away from the reference point?
2 answers
t=3
d(t)= 12
so you have 12= 12 + 3t -t^2
minus 12 from both side
0= -t^2 + 3t factor to get
(-t +3 and t + 0) put them equal to zero and solve for t= 0 and 3
your answer is 3 because zero seconds mean no distance
so you have 12= 12 + 3t -t^2
minus 12 from both side
0= -t^2 + 3t factor to get
(-t +3 and t + 0) put them equal to zero and solve for t= 0 and 3
your answer is 3 because zero seconds mean no distance
1 answer