The question is the roundtrip flight from Edmonton to Toronto covers a distance of 5400km in 7.6 hours. The speed from Edmonton to Toronto was 95 km/h faster than the speed from Toronto to Edmonton. What was the speed of the plane from Toronto to Edmonton?
If i set up my equation as
edmonton to toronto + edmonto back =7.6 hours
5400/x + 5400/95x = 7.6
513000+5400 = 722 x
718=x
Not sure that this is right?
The round trip is 5400, so each half is only 2700 km. So, I get
2700/x + 2700/(95+x) = 7.6
x = 666 km/hr
To solve this problem, let's break it down step by step:
Step 1: Define the variables
Let's define:
- Speed from Toronto to Edmonton as y km/h
- Speed from Edmonton to Toronto as y + 95 km/h (since it's 95 km/h faster than the speed from Toronto to Edmonton)
Step 2: Calculate the time taken for each leg of the trip
The distance from Edmonton to Toronto is 5400 km. Therefore, the time taken from Edmonton to Toronto is:
Time (Edmonton to Toronto) = Distance / Speed = 5400 / (y + 95)
The distance from Toronto to Edmonton is also 5400 km. Therefore, the time taken from Toronto to Edmonton is:
Time (Toronto to Edmonton) = Distance / Speed = 5400 / y
Step 3: Set up the equation
The total time taken for the roundtrip is given as 7.6 hours. Therefore, the equation becomes:
Time (Edmonton to Toronto) + Time (Toronto to Edmonton) = 7.6
5400 / (y + 95) + 5400 / y = 7.6
Step 4: Solve the equation
To solve the equation, we need to find a common denominator. In this case, it is y(y + 95). So, multiplying both sides of the equation by y(y + 95), we get:
5400y + 5400(y + 95) = 7.6y(y + 95)
Expanding and simplifying:
5400y + 5400y + 513000 = 7.6y^2 + 722y
Combining like terms and moving all terms to one side:
0 = 7.6y^2 + 722y - 10800y - 513000
Simplifying further by dividing all terms by 7.6:
0 = y^2 + 95y - 67500
Step 5: Solve the quadratic equation
To solve the quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring may not be straightforward, so let's use the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values:
y = (-95 ± √(95^2 - 4 * 1 * (-67500))) / (2 * 1)
Simplifying further:
y = (-95 ± √(9025 + 270000)) / 2
y = (-95 ± √279025) / 2
Taking the square root:
y ≈ (-95 ± 527.95) / 2
This gives us two possible solutions:
y ≈ (-95 + 527.95) / 2 ≈ 216.97 km/h
y ≈ (-95 - 527.95) / 2 ≈ -311.97 km/h
Since speed cannot be negative, we can conclude that the speed of the plane from Toronto to Edmonton is approximately 216.97 km/h.