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# Physics MCAT 69 Help?

Question 1

A particle has an initial horizontal velocity of 1.8 m/s and an initial upward velocity of 4.7 m/s. It is then given a horizontal accelera- tion of 2.3 m/s2 and a downward acceleration of 1.6 m/s2.

What is its speed after 1.9 s? Answer in units of m/s

Question 2

What is the direction of its velocity at this time with respect to the horizontal? Answer between −180◦ and +180◦.

Answer in units of ◦

### 1 Answer

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- helloLv 69 years agoFavorite Answer
Vf = Vi + at

Vfx = 1.8 + 2.3 x 1.9 work it out

Vfy = 4.7 + (-1.6 x 1.9) work it out.

V^2 = Vfx^2 +Vfy^2 solve for V

at theta = arctan(Vfy/Vfx) use degree mode.

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