Physics problem that confuses me.?

Question

A sailboat is traveling east at 6.0 m/s. A sudden gust of wind gives the boat an acceleration = (0.75 m/s2, 40° north of east). What are the boat's speed and direction 6.0 s later when the gust subsides?


I tried doing vf=vi + a delta t, but evidently that doesn't work. And I thought since the direction became 40 degrees, that's how it would stay, but that didn't work either. Any ideas?

Answer 1

Me too.
You are correct. Treat velocity as it should be treated as a vector quality.

V= at
V(east)= V cos(40)
V(north)= V sin(40)

Vt (east) = V0 + V(east)

Vt (east) = 6.0 + 0.75 x 6.0 cos(40)
Vt (east) = 9.45 m/s
V(north) = V sin(40) = .75 x 6.0 sin(40)
V(north) = 2.9 m/s

Answer 2

Sounds like before the gust, the boat was not accelerating but rather was at a constant velocity 6 m/s due east.

After the gust, there is acceleration which will change the velocity. Assuming the acceleration stays contant over the 6 seconds, after 6 seconds the magnitude of the velocity should be:

6 seconds * 0.75 m/sec-sec = 4.5 m/sec

Now the direction of that new velocity is not due east, but rather it has a northern component. So how much of the 4.5 m/sec velocity adds to the due east 6 m/sec?

4.5 * sin40 = 4.5 * 0.64 = 2.9 m/sec...add this to 6 and get 8.9m/sec due east

How much of the 4.5 m/sec component is due north?

4.5 * cos40 = 4.5 *0.77 = 3.5 msec due north

Does this match the answer in the back of your book ? :)

Answer 3

The neat thing about this stuff is that you CAN figure it out. This problem needs a divide and conquer approach: divide the acceleration and velocity up into N-S and E-W components figure the final v for each and put it back together again.
To start there is no N-S velocity and E-W velocity is 6 m/s
The acc. is 40° and has a mag of 0.75 sin(40°) & cos(40°) for the N-S and E-W accl. (draw a diagram to make sure it matches up roughly with the numbers you get). use those a components to figure the final v's (N-S v and E-W v)
then you can put them back together with hyp²=opp²+adj² and
tan(Θ) = opp/adj (or is it vice versa?) for the angle....angel..

Answer 4

You have right equation but must use it twice (x- and y- components) :

Vfx = Vox +axt
Vfy = V0y +ayt

with
Vox=6
Voy=0
ax=.75cos40
ay=.75sin40
t=6

Solve for Vfx and Vfy. Then use
Vf=sqrt(Vfx^2 + Vfy^2)
angle=arctan(Vfy/Vfx)