Calculating Speed and Direction of Sailboat After Gust of Wind

[Kster]

A sailboat is traveling east at 5m/s . A sudden gust of wind gives the boat an acceleration=.80 m/s^2, (40 degrees north of east).

A. What is the boat's speed 6 seconds later when the gust subsides?
________m/s

B. What is the boat's direction 6 seconds later when the gust subsides?
________degrees north of east.


My attempts:
v = v0 + at
v = (0) + (0.80m/s^2)* (6sec)
v = 4.8m/s

A^2+ B^2 = C^2
(4.8)^2 + (5.0)^2 = 48.04
C= 6.9 m/s
_____________________

tan-1 (4.8/5.0) = 44°


Please tell me what I did wrong? I am so stuck :(

 

[Doc Al]

A^2+ B^2 = C^2
(4.8)^2 + (5.0)^2 = 48.04
C= 6.9 m/s

This assumes that the acceleration is perpendicular to the original velocity (east). But it's not: the acceleration is 40 degrees north of east.

Hint: Break the acceleration (and the velocity) into components (east and north). Find the final velocity component east and the final velocity component north, then you can combine them to find the final speed.

Another approach is to find the change in velocity along the direction of the acceleration, which will be some vector 40 degrees north of east. Then just add that vector to the initial velocity vector and find the new magnitude.