- #1

Ursa

- 11

- 2

- 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##

$$6.0t^2+5.0t=13$$

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##