tourist travels 1500 miles using two planes. The second plane averages 50 miles per hour faster than the first plane. The tourist uses the slower plane for the first 500 and the faster plane for the next 1000 miles. The total flying time is 6.5 hours. What is the speed of the first plane?

V₁ = 200 mi/h

Explanation:

To do this, we need to analyze the given data.

We have a total distance and time of 1500 mi and 6.5 h. We also know that the second plane goes 50 mi/h faster than the first plane, (V₂ = V₁ + 50). The firsts 500 miles were on the first plane and the remaining 1000 mi with the second plane.

Using the expression of distance, and assuming a constant speed in both planes we have:

d = V * t   (1)

Replacing this expression for both planes we have:

d₁ = V₁ * t₁

d₂ = (V₁+50) * t₂

WE are missing several data, because we do not know the times for each plane, but we can have a clue.

The total time of flight was 6.5 h, this is equivalent to just sum both times of both planes, so:

t = t₁ + t₂   (3)

From the expressions of distance for both planes, we can solve for time, and then replace in the above expression of the total time. In that way we can solve for the speed of the plane. Let's solve for both times in the planes:

t₁ = d₁ / V₁ = 500/V₁    (4)

t₂ = d₂ / V₁+50 = 1000/ V₁ + 50   (5)

Now, all we have to do is replace (4) and (5) into (3) and solve for V₁:

6.5 = [(500/V₁) + (1000/V₁ + 50)

6.5 = [500*(V₁+50) + 1000V₁ / V₁(V₁+50)]

6.5 = [500V₁ + 25000 + 1000V₁ / V₁² + 50V₁

6.5(V₁² + 50V₁) = 1500V₁ + 25000

6.5V₁² + 325V₁ - 1500V₁ - 25000 = 0

6.5V₁² - 1175V₁ - 25000 = 0

From here, we use the quadratic equation to solve for V₁:

X = 1175 ± √(1175)² + 4 * 6.5 * 25000 / 2 * 6.5

X = 1175 ± √1380625 + 650000 / 13

X = 1175 ± 1425 / 13

X₁ = 1175 + 1425 / 13 = 200

X₂ = 1175 - 1425 / 13 = -19.23

So, it's very clear that the speed cannot be negative, so we stick with the first value. Therefore the speed pf the first plane is:

Hope this helps

a electromagnetic waves

true is the correct answer

hope i