2D acceleration

[F.B]

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??

[BobbyJoeCool]

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?

[mezarashi]

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.

[F.B]

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.

[BobbyJoeCool]

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]

[F.B]

The answer in the book says 3.45(km/h)/s [52.7 West of South]

[BobbyJoeCool]

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?