A plane flies directly from city A to B, which are separated by 2300mi. From A to B, the place flies into a 65mi/hr headwind. ON the return trip from A to B, the wind velocity is unchanged. The trip from B to A takes 44 min less than the trip from A to B. What is the airspeed (assumed constant) of the plane?
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To find the airspeed of the plane, we need to break down the information given and use the concept of relative velocity.
Let's assume the airspeed of the plane is V mph, and the wind speed is W mph.
On the trip from A to B:
- The plane is flying *against* a 65 mph headwind, so the effective speed of the plane is (V - 65) mph.
On the return trip from B to A:
- The wind speed remains unchanged at 65 mph, so the plane is flying *with* a 65 mph tailwind. Therefore, the effective speed of the plane is (V + 65) mph.
We know that the distance between city A and B is 2300 miles.
Using the formula: Distance = Speed × Time, we can calculate the times for each journey.
- For the trip from A to B, the time is given by: 2300 = (V - 65) × T1, where T1 is the time taken.
- For the trip from B to A, the time is given by: 2300 = (V + 65) × T2, where T2 is the time taken.
Now we are given that the return trip (B to A) took 44 minutes less than the trip from A to B.
Thus, T2 = T1 - 44/60 hours (converting 44 minutes to hours).
Substituting this value into the equation for the trip from B to A, we get: 2300 = (V + 65) × (T1 - 44/60).
Now we have two equations:
1. 2300 = (V - 65) × T1
2. 2300 = (V + 65) × (T1 - 44/60)
To solve these equations, we can use substitution or elimination methods. Let's use the substitution method:
Rearrange equation 1 to solve for T1:
T1 = 2300 / (V - 65)
Substitute the value of T1 in equation 2:
2300 = (V + 65) × (2300 / (V - 65) - 44/60)
Now, we can simplify and solve this equation for V, the airspeed of the plane.