# Velocity Vectors

Posted by Professor Cram in Weblog

Dear Professor Cram:

An executive flew his corporate jet to a meeting in a city 1500 kilometers away. After traveling the same amount of time on return flight, the pilot mentioned that they still had 300 km to go. The air speed of the plane was 600 km/hr for both legs. Assuming that the wind direction was parallel to the flight path and constant all day, how fast was the wind blowing?

Ekta, Middlesex Community College

Ekta, thank you for this physics question involving vectors. Let's see how we can tackle it.

I'm assuming that the constant 600 km/hr was the instrument output of the plane, not the actual as-observed-from-the-ground speed. Thus, the trip out was clocked at 600 km/hr plus the tailwind, while the return trip was at 600 km/hr minus the tail wind.

In this problem, the actual speed of the plane is a vector combination of plane speed and wind speed. On the first trip, both speeds are working together so we have a vector sum; on the return trip, the vectors are working against each other. (Reality check — because the plane made it there and partway back, that means the plane speed vector is larger than the wind speed vector.)

According to the question, during the first leg the travel time can be expressed as:

- time = distance/speed, or
- time = 1500/(600 + x)

where x is the speed of the wind. On the return leg, the same amount of time elapsed but the plane still had 300 km to go.

- time = distance/speed, or
- time = (1500-300)/(600 – x), or
- time = 1200/(600 – x)

Since these two times are equal, we can set both equations equal to each other and solve for x:

- 1500/(600 + x) = 1200/(600 – x), or
- (1500)(600 – x) = (1200)(600 + x), or
- 900,000 – 1500x = 720,000 + 1200x, or
- 900,000 – 720,000 = 1200x + 1500x, or
- 180,000 = 2700x, or
- x = 66 2/3

The wind was blowing at 66 2/3 km/hr.

Good studying,

Professor Cram

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