Projectile Motion wind acceleration

[soul5]

Homework Statement

A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration a = (6i + 7j)m/s^2 . At time t = 0, the velocity is (3i)m/s . What are the (a) magnitude and (b) angle of its velocity when it has been displaced by 12.0 m parallel to the x axis?


i= acceleration in x component , j = acceleration in y component

The Attempt at a Solution

vx^2 = (3m/s)^2 + 2 (6m/s^2) (12m/s)


vx^2= 9m/s + 144m/s

vx = 12.36


I did that and got velocity wrong what do I do?

 

[LowlyPion]

You're only half way home.

What's the Vy?

Just because it has no initial velocity in j doesn't mean that it isn't swept by the wind in the j direction.

 

[soul5]

You're only half way home.

What's the Vy?

Just because it has no initial velocity in j doesn't mean that it isn't swept by the wind in the j direction.

So once I find Vy I use Vx^2 + Vy^2 = V^2?

 

[LowlyPion]

So once I find Vy I use Vx^2 + Vy^2 = V^2?

That would be the way to do it.

And tan^-1(Vy/Vx) would give the angle to positive x.

 

[soul5]

how do I find vy?

 

[LowlyPion]

Along Vx how long did it take to cover the 12m?

Use that time to determine Vy.

Vy = a*t