Answer: the two times such that the car is 5 miles away from the checkpoint are 1.5 hours and 1.83 hours.
Step-by-step explanation:
We know that:
At the beginning, at t = 0h, the car is 50 miles away from the checkpoint (let's define the checkpoint as the position 0mi).
And the speed of the car is 30mi/h.
Then the movement equation for the car will be:
d(t) = 50mi - 30mi/h*t
Then at t = 0h, we will have:
d(t) = 50mi - 30m/h*0h = 50mi
So at the beggining, the distance to the checkpoint is 50 miles, so it is consistent with our case.
Now we want to find the times such that the car is 5 miles away from the checkpoint.
These times are when:
d(t1) = 5mi
d(t2) = -5mi
Then let's solve these two equations:
d(t1) = 5mi = 50mi - 30mi/h*t1
-45mi = -30mi/h*t1
(45/30) h = t1 = 1.5 hours.
And the other time is:
d(t2) = -5mi = 50mi - 30mi/h*t2
-55mi = -30mi/h*t2
(55/30) h = t2 = 1.83 hours.
Then the two times such that the car is 5 miles away from the checkpoint are 1.5 hours and 1.83 hours.
