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?
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
To determine when the car is 12 units away from the reference point, we need to solve the equation d(t) = 12. Let's substitute the equation given for d(t) into the equation:
12 + 3t - t^2 = 12
By rearranging the equation, we get:
t^2 - 3t = 0
Now we can factor out t:
t(t - 3) = 0
To find the values of t, we set each factor equal to zero:
t = 0 or t - 3 = 0
From the first equation, t = 0, we can conclude that the car is at the reference point when t = 0 seconds.
From the second equation, t - 3 = 0, we find that t = 3. This means the car is also 12 units away from the reference point after 3 seconds.
Therefore, the car is 12 units away from the reference point at t = 0 seconds and t = 3 seconds.