Starting from 50 miles away, a car drives toward a speed checkpoint and then passes it. The car travels at a constant rate of 30 miles per hour.
The distance of the car from the checkpoint is given by d=150-3001
At what times is the car 5 miles from the checkpoint?
Plz show steps

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.

answer: it is 1: 47 pm

step-by-step explanation:

first you will have to convert 3/4 into a decimal

which is .75

then you will have to divide 1.5 by .75

which is 2

no you have to divide