# physcis

a particle moving at a velocity of 6 m/s in the positive x direction is given an acceleration of 3.9m/s^2 in the positive Y direction for 3.4s. what is the final speed of the particle? answer in units of m/s

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1. The x component of velocity remains
Vx = 6 m/s.
The y component of velocity increases with time, t, and is
Vy = 3.9 t. After 3.4 seconds
Vy(@ t=3.4s) = 13.3 m/s

The final speed (the magnitude of velocity) is sqrt(Vx^2 + Vy^2)

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posted by drwls
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posted by gago

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