Question

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?

Answer 1

you know vectors?

Answer 2

no :/

Answer 3

what's the answer?

Answer 4

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.

Answer 5

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

Answer 6

5.1108 but what's the equation of motion

Answer 7

then use, v=u+at

Answer 8

u is initial speed (0)

Answer 9

I got 37.81 but that seems like a lot?

Answer 10

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.

Answer 11

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

Answer 12

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.

Answer 13

Thanks for your help :)

Answer 14

yeah, np