Aparticle travels to the right at a constant rate of 6.8 m/s. it suddenly is given a vertical acceleration of 3.4 m/s 2 for 4.3 s. what is its direction of travel after the acceleration with respect to the horizontal? answer between −180◦ and +180◦ .

65.05°

Explanation:

Hello!

In order to know the direction of travel after the acceleration we must first know the velocity after the acceleration.

The horizontal component of velocity remains the same, but the vertical component will be given as the product of the (constant) acceleration times the total time.

Vx = 6.8 m/s

Vy = (3.4)*(4.3) m/s = 14.62 m/s

Then the angle of the motion of the particle with respect to the horizontal is given by:

Ф = arctan(Vy/Vx) =arctan(14.62/6.8) = 65.05°

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