# 2-Dimensional Kinetic Motion

1. Oct 14, 2007

### ideefixem

1. The problem statement, all variables and given/known data
A sailboat is traveling east at 5.0 m/s. A sudden gust of wind gives the boat an acceleration of 0.80 m/s^2 at 40 degrees north of east. What are the boat's speed and direction 6.0 s later when the gust subsides?

2. Relevant equations
xf=xi+vix(delta t)+1/2ax(delta t)^2
vfx=vix+ax(delta t)

3. The attempt at a solution
This is really giving me a tough time.

I started by trying to determine the velocity of both the x and y values.

vfx=vix+ax(delta t)
vfx=5m/s+(0)(6)
vfx=5m/s

vfy=5m/s+(.8)(6)
vfy=9.8 m/s

v=sqrt(vx^2+vy^2)
v=sqrt(5^2+9.8^2)
v=11.002 m/s

I really need help determining the position after 6s.

2. Oct 14, 2007

### learningphysics

You don't have ax and ay right...

The acceleration is 0.80m/s^2 at 40 degrees north of east... use trig to get ax and ay...

Then as you did

vfx = vix + (delta t)ax

vfy = viy + (delta t)ay

And just use your formula

xf=xi+vix(delta t)+1/2ax(delta t)^2

for the position... xi = 0. and use the corresponding equation in the y-direction.

3. Oct 15, 2007

### ideefixem

Thank you very much for your help...
I think this is correct... could you please review it for me?

cos40(.8)= ax= 0.613 m/s^2
sin40(.8)= ay= 0.514 m/s^2

Vfx= 5+(0.613)(6)= 8.68 m/s
Vfy= 5+(0.514)(6)= 8.08 m/s
V= sqrt(8.68^2+8.08^2)= 11.86 m/s

xf=0+5(6)+.5(.613)(6)^2= 41.034 m
yf=0+5(6)+.5(.514)(6)^2= 31.542 m

tan-1(31.542/41.034)= 37.55 degrees NE

4. Oct 15, 2007

### learningphysics

Viy = 0. so

Vfy= 0+(0.514)(6)= 3.084 m/s

so V = sqrt(8.68^2+3.084^2)= 9.21m/s.

and the direction is just arctan(3.084/8.68) = 19.56 degrees north of east.

Reading over the question, i don't think you need position... just speed and direction...