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Ahat [919]
3 years ago
14

Starting from 50 miles away, a car drives toward a speed checkpoint and then passes it. The car

Mathematics
1 answer:
padilas [110]3 years ago
8 0

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.

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Mary drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 7 hours. When Mary drove h
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There are two separate equations for this question that use the same variable.

Mary took 7 hours to complete a trip with traffic. This would use the following expression:

7x

Mary took 4 hours to complete the same trip without traffic. The average rate of speed was 27 mph faster than the trip with traffic. This uses the following expression:

4(x+27)

Both trips had the same distance. From the information provided, we can set up the following equation:

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