A plane flew from Calgary to Toronto, a flying distance of 2720 km. On the return journey, due to a strong head wind, the average flying speed was 80km/hr slower than on the outward journey. The time taken for the return journey was 40 minutes more than for the outward journey. Calculate the time taken for the journey from Calgary to Toronto. Calculate the time time taken for the whole trip. Calculate the average speed of the plane from Toronto to Calgary.
40 min = .667 hour
ground plane speed in still air = v
speed against the wind = (v-80)
Tct = 2720 / v
Ttc = 2720 / (v-80)
Ttc - Tct = .667 hour = 2720 / (v-80) - 2720 /v
.667 (v-80)(v) = 2720 [ v -v+80 ]
.667 v^2 - 53.3 v - 217600 = 0
v = 613 km/h or -533 km/h, use the positive root, 613 km/h
To calculate the time taken for the journey from Calgary to Toronto, we can use the distance and average speed. Let's assume the average speed of the plane on the outward journey is x km/hr.
The formula for calculating time is:
Time = Distance / Speed
For the outward journey:
Time_1 = 2720 km / x km/hr
For the return journey, the average flying speed was 80 km/hr slower than the outward journey, so the average speed will be (x - 80) km/hr.
Let's calculate the time for the return journey:
Time_2 = 2720 km / (x - 80) km/hr
We are given that the time taken for the return journey was 40 minutes more than the outward journey. Since time is measured in hours, we need to convert 40 minutes to hours:
40 minutes = 40/60 = 2/3 hours
Now we can set up an equation using the given information:
Time_2 = Time_1 + 2/3
Substituting the values of Time_1 and Time_2, we get:
2720 / (x - 80) = 2720 / x + 2/3
Next, we can solve this equation step by step.
Multiply both sides of the equation by x(x - 80) to eliminate the denominators:
2720x = 2720(x - 80) + (2/3)(x)(x - 80)
Simplifying the equation:
2720x = 2720x - 217600 + (2/3)(x^2 - 80x)
Expanding the second term on the right side of the equation:
2720x = 2720x - 217600 + (2/3)(x^2) - (2/3)(80x)
Combining like terms:
2720x = 2720x - 217600 + (2/3)(x^2) - (160/3)x
Rearranging the terms:
0 = (2/3)(x^2) - (160/3)x - 217600
Multiplying both sides by 3 to eliminate the fractions:
0 = 2x^2 - 160x - 652800
Now we can solve this quadratic equation. Either by factoring or using the quadratic formula:
By factoring, we look for two numbers whose sum is -160 and product is -652800. After some calculations, we find that the factors are:
(x - 680)(2x + 960) = 0
This gives two possible solutions:
1) x - 680 = 0, which means x = 680
2) 2x + 960 = 0, which means x = -480 (ignore this negative value since it does not make sense in this context)
Therefore, the average speed of the plane on the outward journey is 680 km/hr.
Now, let's calculate the time taken for the journey from Calgary to Toronto:
Time_1 = 2720 km / 680 km/hr
Time_1 = 4 hours
The time taken for the whole trip is the sum of the time for the outward journey and the time for the return journey:
Total Time = Time_1 + Time_2
Total Time = 4 hours + (Time_1 + 2/3) hours
Total Time = 4 hours + 4 hours + 2/3 hours
Total Time = 8 hours + 2/3 hours
To combine the whole hours and the fraction, let's write 2/3 as 40/60 (which is equal to 2/3):
Total Time = 8 hours + 40/60 hours
Simplifying the fraction:
Total Time = 8 hours + 2/3 hours
Total Time = 8 2/3 hours
Therefore, the time taken for the whole trip is 8 hours and 40 minutes.
Finally, let's calculate the average speed of the plane from Toronto to Calgary. The distance for the return journey is the same as the outbound journey, which is 2720 km. The time for the return journey is 8 2/3 hours:
Average Speed = Distance / Time
Average Speed = 2720 km / (8 2/3) hours
To convert the mixed fraction to an improper fraction:
Average Speed = 2720 km / (26/3) hours
To divide a fraction by a fraction, we multiply by the reciprocal of the divisor:
Average Speed = 2720 km * (3/26) hours
Simplifying:
Average Speed = 520 km/26
Average Speed = 20 km/hr
Therefore, the average speed of the plane from Toronto to Calgary is 20 km/hr.
To calculate the time taken for the journey from Calgary to Toronto, we need to find the speed and use the distance formula: time = distance / speed.
Let's assume the speed of the plane on the outward journey is S km/hr. We know that the distance is 2720 km.
So, the time taken for the outward journey is: time_outward = 2720 / S.
On the return journey, the plane's average speed was 80 km/hr slower than the outward journey. Therefore, the speed on the return journey is (S - 80) km/hr. The distance traveled is the same, 2720 km.
So, the time taken for the return journey is: time_return = 2720 / (S - 80).
We also know that the time taken for the return journey is 40 minutes longer than the outward journey. Since 1 hour is equal to 60 minutes, we can convert the 40 minutes to hours by dividing it by 60.
So, the time_return = time_outward + 40/60.
Now we can set up an equation to solve for the time taken for the outward journey:
2720 / S = 2720 / (S - 80) + 40/60.
To solve this equation, we need to simplify and solve for S. Let's start by multiplying both sides of the equation by S(S - 80) to eliminate the denominators:
2720(S - 80) = 2720S + 2/3(S)(S - 80).
We can simplify further by expanding the terms:
2720S - 217600 = 2720S + 2/3(S^2 - 80S).
Cancel out the common terms on both sides:
-217600 = 2/3(S^2 - 80S).
Multiplying both sides by 3 to eliminate the fraction, we get:
-652800 = 2(S^2 - 80S).
Expanding the terms:
-652800 = 2S^2 - 160S.
Now, rearrange the equation to form a quadratic equation by moving all the terms to one side:
2S^2 - 160S - 652800 = 0.
We can now use the quadratic formula to solve for S:
S = (-b ± √(b^2 - 4ac)) / (2a).
In our case, a = 2, b = -160, and c = -652800. Plug these values into the formula:
S = (-(-160) ± √((-160)^2 - 4(2)(-652800))) / (2(2)).
After simplifying, we get:
S = (160 ± √(25600 + 5222400)) / 4.
Further simplifying:
S = (160 ± √(5248000)) / 4.
The positive value of S will give us the average speed of the plane from Calgary to Toronto. Let's calculate it:
S = (160 + √5248000) / 4.
S ≈ 134.96 km/hr (rounded to two decimal places).
Now that we have the speed, we can calculate the time taken for the journey from Calgary to Toronto:
time_outward = 2720 / S ≈ 2720 / 134.96 ≈ 20.16 hours (rounded to two decimal places).
To calculate the time taken for the whole trip, we can find the sum of the time taken for the outward journey and the return journey:
time_whole_trip = time_outward + time_return ≈ 20.16 + (time_outward + 40/60).
Now, let's calculate the time taken for the return journey:
time_return = 2720 / (S - 80) ≈ 2720 / (134.96 - 80) ≈ 2720 / 54.96 ≈ 49.49 hours (rounded to two decimal places).
Plugging in the values:
time_whole_trip ≈ 20.16 + (49.49 + 40/60) ≈ 20.16 + 49.82 ≈ 70.98 hours (rounded to two decimal places).
Therefore, the time taken for the journey from Calgary to Toronto is approximately 20.16 hours, the time taken for the whole trip is approximately 70.98 hours, and the average speed of the plane from Toronto to Calgary is approximately 134.96 km/hr.