1 / 4

Velocity Triangles and Relative Velocity

Velocity Triangles and Relative Velocity. A typical problem :

beverly-ramos
Download Presentation

Velocity Triangles and Relative Velocity

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Velocity Triangles and Relative Velocity A typical problem: A helicopter can fly at 200 km h1 in still air. It has to travel from an airport to a hospital. The hospital is 600 km from the airport on a bearing of 250. The wind speed is 80 km h1 and it is blowing from the south west. Work out the course that the helicopter must fly on and the time that it takes to get to the hospital.

  2. First step: construct the velocity triangle Two things to look out for: • the resultant velocity must be in the direction of the hospital from the airport, i.e. the path that the helicopter has to fly along. • the direction of the vectors in the velocity triangle must follow this rule: resultant velocity  initial velocity  change in velocity e.g. Speed in still water or still air. This will be in the direction of the course that has to be set. This will usually be either the wind’s velocity or the current’s velocity.

  3. This is the resultant velocity North Airport 250 Hospital North 80 200 45 Helicopter’s speed in still air, in the direction of the course it must fly on. Wind blowing FROM south west

  4. Find the size of an internal angle of the triangle, using basic geometry - the north lines [dashed] are parallel. 110  70 135 • Use Sine Rule to find  and hence : • sin   sin 155 • 80 200 •   97 [1 d.p.] and   153 155 80 200  45 • Use Sine Rule again to find unknown side length [resultant speed] : • 80 speed • sin  sin  • speed  125 km h1 [3 s.f.] Hence the helicopter must fly on a bearing of 240  [to nearest degree] and the time taken is 4 hours 48 minutes.

More Related