Problem #3
Science and medicine. Adriana took 2 h longer to drive 360 mi on the first day of a trip than she took to drive 270 mi on the second day. If her speed was the same on both days, what was the driving time each day?
my answer: Her driving time was 8h for her first day and 6 hours the second day.
Problem#4
Science and medicine. A plane flies 720 mi against a steady 30-mi/h headwind and then returns to the same point with the wind. If the entire trip takes 10 h, what is the plane’s speed in still air?
My answer: The speed of the plane in still air is 150mi/h
both correct.
A pilot flies 750 miles west (with the wind). On the return flight (against the wind) following the same flight path, she travels only 600 miles in the same time. Find the speed of the plane. Wind is 15 mph.
To solve Problem #3, we need to set up a system of equations. Let's denote the driving time on the first day as x hours and the driving time on the second day as y hours.
We know that Adriana took 2 hours longer to drive 360 miles on the first day than to drive 270 miles on the second day. This can be represented by the equation:
x = y + 2
We also know that her speed was the same on both days. Since speed = distance/time, we can set up the following equation:
360/x = 270/y
To solve this system of equations, we can substitute the value of x from the first equation into the second equation:
360/(y + 2) = 270/y
To simplify this equation, we can cross-multiply:
360y = 270(y + 2)
Expanding the equation:
360y = 270y + 540
Simplifying further:
90y = 540
Dividing both sides by 90:
y = 6
Now that we know y is 6, we can plug it back into the first equation to find x:
x = y + 2 = 6 + 2 = 8
Therefore, Adriana's driving time was 8 hours on the first day and 6 hours on the second day.
To solve Problem #4, we can use the formula for time:
time = distance/speed
Let's denote the plane's speed in still air as s mi/h.
When the plane flies against the headwind, its effective speed is reduced. The time taken for this leg of the journey can be calculated using:
720/(s - 30) hours
When the plane flies with the wind, its effective speed is increased. The time taken for this leg of the journey can be calculated using:
720/(s + 30) hours
Now, we know that the entire trip took 10 hours. So, we can set up the following equation:
720/(s - 30) + 720/(s + 30) = 10
To simplify this equation, we can first find a common denominator:
(720(s + 30) + 720(s - 30))/((s - 30)(s + 30)) = 10
Expanding the numerator:
(720s + 21600 + 720s - 21600)/(s^2 - 900) = 10
Simplifying further:
(1440s)/(s^2 - 900) = 10
Multiplying both sides by (s^2 - 900):
1440s = 10(s^2 - 900)
Expanding the equation:
1440s = 10s^2 - 9000
Rearranging the equation:
10s^2 - 1440s - 9000 = 0
Now we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Once we find the values of s, we can determine the speed of the plane in still air.
In this case, the quadratic equation cannot be easily factored. So, let's use the quadratic formula:
s = (-b ± √(b^2 - 4ac))/2a
Here, a = 10, b = -1440, and c = -9000. Substituting the values into the quadratic formula, we get:
s = (-(-1440) ± √((-1440)^2 - 4(10)(-9000)))/(2(10))
s = (1440 ± √(2073600 + 360000))/(20)
s = (1440 ± √(2433600))/(20)
Calculating the square root:
s = (1440 ± 1560)/(20)
s = (1440 + 1560)/(20) or s = (1440 - 1560)/(20)
Simplifying further:
s = 300/20 or s = -120/20
s = 15 or s = -6
Since speed cannot be negative, the speed of the plane in still air is 15 mi/h.
Therefore, the correct answer to Problem #4 is that the plane's speed in still air is 15 mi/h.