## RavenLynette Group Title A particle at rest undergoes an acceleration of 2.6 m/s^2 to the right and 4.4 m/s^2 up What is its speed after 7.4 s? 11 months ago 11 months ago

1. shubham.bagrecha Group Title

you know vectors?

2. RavenLynette Group Title

no :/

3. shubham.bagrecha Group Title

|dw:1379213971917:dw|

4. shubham.bagrecha Group Title

5. ybarrap Group Title

compute the magnitude of the vector (i.e. Pythagoras theorem) as suggested by @shubham.bagrecha . Then use your equations of motion to determine speed using this magnitude.

6. shubham.bagrecha Group Title

net acceration ,$a=\sqrt{4.4^2+2.6^2}$

7. RavenLynette Group Title

5.1108 but what's the equation of motion

8. shubham.bagrecha Group Title

then use, v=u+at

9. ybarrap Group Title

u is initial speed (0)

10. RavenLynette Group Title

I got 37.81 but that seems like a lot?

11. ybarrap Group Title

5 m/s^2 is about 1/2 g. If you fell for 7 seconds at 1 g your speed would be 72 m/s, so 37.8 doesn't seem to far off what you would expect given that acceleration.

12. RavenLynette Group Title

Oh wow well thanks :) could I use cosine to find the direction between -180 and 180 degrees?

13. ybarrap Group Title

You could use $$\arctan(4.4/2.6)$$ to find the angle between the direction of motion and the horizontal, which in this case is about 60$$^\circ$$. But, it's not necessary to solve this problem.

14. RavenLynette Group Title