Answer:
>34.88km/h
Explanation:
Because we have two unknown values in this question (speed of train and speed of car), we will need two equations to solve it.
Firstly, we can make a big equation for time in terms of the vehicle speeds. Time = distance / speed, so if we use t for the train speed and c for the car speed, we get this equation for speed:
30/c + 90/t = 2.5
Here the 30 is the distance traveled by the car in km and the 90 is the distance traveled by the train in km (120 - 30 = 90).
We also know that the train speed was 20km/h faster than the car speed so our second equation is:
c + 20 = t
So now we can substitute this in the big equation so we only have one variable to solve for:
30/c + 90/(c+20) = 2.5
Now to solve this, this first step is to get rid of the fractions, which we can do by multiplying all the terms by the lowest common factor of the two denominators which is c(c+20):
(30/c)*c(c+20) + (90/c+20)*c(c+20) = 2.5
This simplifies to:
120c+600=2.5c^2+50c
and then into the quadratic equation:
2.5c^2-70c-600=0
This can be solved with the quadratic formula or a solver/calculator, and yields the solutions:
c=−6.88, 34.88 (both to 2dp)
We can eliminate the negative solution because speed cannot be negative meaning that Rosea must drive her car at the speed of at least/greater than 34.88km/h if she wants the total trip to be less than 2.5 hours.
Hope this helped!