A pilot must fly his plane to a town which is 200km from his starting point in a direction 30degrees N of E. He must make the trip in 1/2 h. An 80km/h wind blows in a direction 30degrees E of S. Find the speed of the plane relative to air.
I found the speed of the plane relative to the ground is 400km/h.
The answer for the speed of the plane relative to the air is 408 km/h. Can someone please explain HOW you would get that.
Check my reply to this yesterday at 12:27pm
Thanks! I understood it now. Do you know what the angle between then pVw and wVg vector would be?
Also can you please help me with the continuation of this problem.
Suppose the pilot disregarded the wind and had flown in a direction of 30 degrees N of E. Where relative to his destination would he be when he thought he should be at the destination?
I don't know what you mean by "the angle between the pVw and wVg vector".
For the second part:
The draw the plane vector which is the same as the desired results. At the end of the plane vector, draw the wind vector. That is where the plane will end up. So, for a 1/2 hour flight:
(80 km/h) * (hour/2) at 30 degrees E of S.
or 40km at 30 degrees E of S of the desired destination.
To find the speed of the plane relative to the air, we need to consider the effect of the wind on the plane's motion. Let's break down the problem step by step:
1. Find the plane's velocity relative to the ground:
- The distance between the starting point and the town is 200 km.
- The pilot needs to cover this distance in 1/2 hour.
- Thus, the plane's velocity relative to the ground is given by v_p = 200 km / 0.5 hour = 400 km/h.
2. Decompose the velocity of the wind into north-south (vertical) and east-west (horizontal) components:
- The wind blows in a direction 30 degrees east of south.
- The speed of the wind is given as 80 km/h.
- By using trigonometry, we can calculate the horizontal and vertical components of the wind's velocity. Let's call the horizontal component v_wx and the vertical component v_wy.
- v_wx = 80 km/h * cos(30 degrees) ≈ 69.282 km/h
- v_wy = 80 km/h * sin(30 degrees) ≈ 39.14 km/h
3. Calculate the resultant velocity of the plane relative to the air:
- Since the wind is blowing from east to west, it will have an impact on the plane's velocity.
- The plane's velocity relative to the ground (v_p) is the vector sum of the plane's velocity relative to the air (v_pa) and the wind's velocity (v_w), resulting in the equation v_p = v_pa + v_w.
- Considering the north-south and east-west components separately, we can write two equations:
- 400 km/h (plane's velocity relative to the ground) = v_pax (plane's velocity relative to the air) + v_wx (horizontal component of wind's velocity)
- 0 (since the plane is not flying vertically) = v_pay (vertical component of plane's velocity relative to the air) + v_wy (vertical component of wind's velocity).
- We can rearrange the equations to find v_pax and v_pay:
- v_pax = 400 km/h - v_wx ≈ 400 km/h - 69.282 km/h ≈ 330.718 km/h
- v_pay = -v_wy ≈ -39.14 km/h (here, the negative sign indicates that the plane is flying opposite to the direction of the wind)
- Therefore, the speed of the plane relative to the air is the magnitude of the resultant vector: v_pa = √(v_pax^2 + v_pay^2) ≈ √(330.718 km/h)^2 + (-39.14 km/h)^2 ≈ 408 km/h.
Hence, the speed of the plane relative to the air is approximately 408 km/h.