Pebble accelerated by the wind with a starting velocity

[Ursa]

Homework Statement

A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration $\vec a = (6.0i + 7.0j)m/s^2$. At time t = 0, the velocity is $(5.0i)m/s$ . What are the (a) magnitude and (b) angle of its velocity when it has been displaced by 13.0 m parallel to the x axis?

Relevant Equations

$$x = \frac {-b \pm \sqrt{b^2 -4ac}} {2a}$$
$$V=\sqrt {V_x^2 + V_y^2} $$
$$a_x=a \cos\Theta$$

I first used differentials to find an equation for the displacement.
$$6.0t^2+5.0t=13$$
and using the quadratic formula I got time $t=1.1$
I then got $V_x$ from $6.0t+5.0=v_x=11.6$
and $V_y$ from $7.0t=v_y=7.7$
The I got v from $V=\sqrt {V_x^2 + Vy^2}$
$$\sqrt {11.6^2 + 7.7^2} =13.9 m/s$$

Which I took as my velocity.

from there I used the vector vector competent equation $a_x=a \cos\Theta$ to find the angle

$\cos^{-1}\frac {11.6} {13.9}=\Theta=33.4$

 

[TSny]

I first used differentials to find an equation for the displacement.
$$6.0t^2+5.0t=13$$

The factor of 6.0 in the first term is not correct. Check the standard kinematics equations for displacement with constant acceleration. Otherwise, your method of solution looks good.

 

[haruspex]

There is no need to find the time.
There are five standard SUVAT equations, each involving four of time, displacement, acceleration, initial velocity, final velocity.
Which three of those are you given, and which are you trying to find? Which SUVAT equation uses just those four?