F.B Posted September 20, 2005 Share Posted September 20, 2005 My teacher is driving me crazy he doesnt explain everything. Anyways my question is this: A helicopter travelling horizontally at 155 km/h [East] executes a gradual turn, and after 56.5s flies at 118 km/h [south]. What is the helicopter's average acceleration in kilometres per hour per second. I dont get how we are supposed to do this without angles. how am i supposed to find the components of x and y. Can anyone please help me?? Link to comment Share on other sites More sharing options...
BobbyJoeCool Posted September 20, 2005 Share Posted September 20, 2005 My teacher is driving me crazy he doesnt explain everything. Anyways my question is this: A helicopter travelling horizontally at 155 km/h [East] executes a gradual turn' date=' and after 56.5s flies at 118 km/h [south']. What is the helicopter's average acceleration in kilometres per hour per second. I dont get how we are supposed to do this without angles. how am i supposed to find the components of x and y. Can anyone please help me?? you do have an angle... from east to south=90°. You have a beginning vector, and the ending vector, and the angle between them (90°). Finding the other side is quite easy... [math]v_{a}^2+v_{b}^2=v_{c}^2[/math] (note: this is only true because it's a right angle (right triangle)) Normally it'd be [math]v_{c}^2=v_{a}^2+v_{b}^2-2v_{a}v_{b}cos©[/math] Since cos(90)=0, 2(v)(v)(0)=0 and you're left with the above. [math]v_{c}[/math] is the vector you're solving for. [math]v_{c}=\sqrt{v_{a}^2+v_{b}^2-2v_{a}v_{b}cos©}[/math] input numbers... any help? Link to comment Share on other sites More sharing options...
mezarashi Posted September 20, 2005 Share Posted September 20, 2005 How about, the definition of average acceleration is simply: Final Velocity (can be a vector) - Initial Velocity (vector as well) over the time needed to accomplish this change in velocity. Link to comment Share on other sites More sharing options...
F.B Posted September 21, 2005 Author Share Posted September 21, 2005 i tried pythagorean it doesnt work, unless the answer in the book is wrong. Also using the actual average acceleration equation, it still doesnt work i have no idea whats wrong. I think you need to convert the 155 and the 118 to magnitude and then use pythagorean but i have no idea how because you have no angle. Link to comment Share on other sites More sharing options...
BobbyJoeCool Posted September 21, 2005 Share Posted September 21, 2005 what answer does the book give, and we'll see if we can help you get there... EDIT: also, in the above explanation, you only end up with the change in velocity... you still need to divide it by the time it takes to do this... [math]\frac{v_{c}^2}{s}=a[/math] Link to comment Share on other sites More sharing options...
F.B Posted September 21, 2005 Author Share Posted September 21, 2005 The answer in the book says 3.45(km/h)/s [52.7 West of South] Link to comment Share on other sites More sharing options...
BobbyJoeCool Posted September 21, 2005 Share Posted September 21, 2005 The answer in the book says 3.45(km/h)/s [52.7 West of South] Ok... I think I can answer this now... You need to accelerate 155 km/hr WEST (to cancil your eastwards moment). You also need to accelerate 118 km/hr south, so that you're moving that direction. the resultant vector is those two added, so that you are moveing 118 km/hr south. Try doing it that way... (and remember, you have a 90°*angle ). I see where it's going now... I get, 3.45 km/(h*s) you use pythagoreans and then devide by the time for average acceleration... the angle is just some trig... help much at all? Link to comment Share on other sites More sharing options...
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