# Particle with horizontal velocity and vertical acceleration

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1. Jun 16, 2013

### Robertoalva

1. A particle moving at a velocity of 6.5 m/s in the positive x direction is given an acceleration of 0.9 m/s^2 in the positive y direction for 2.9 s. What is the final speed of the particle?

2. Relevant equations
v= vi +at

3. The attempt at a solution

v= 6.5 + (.9)(2.9)= 9.11 m/s

I don't know if the velocity is right. the velocity is horizontal, and acceleration is vertical, so it makes me doubt.

2. Jun 16, 2013

### Staff: Mentor

The horizontal and vertical components of the velocity are independent; They should be treated as vector components when you "add" them. The final speed will be the magnitude of the resultant vector.

3. Jun 16, 2013

### Robertoalva

so, i have to use the following, vy = viy +ay t and vx= vix + ax t ??

4. Jun 16, 2013

### UVW

That's right. Before, you were using v = vy + axt, which doesn't make sense, since those directions are perpendicular. You'll need to take gneill's advice about what to do from there.

5. Jun 16, 2013

### Robertoalva

vy = viy +ay t = viy + (.9)(2.9)

vx= vix + ax t = (6.5) + ax (2.9)

how do i get the missing values?

6. Jun 16, 2013

### Staff: Mentor

Initially the particle had only a velocity in the x-direction. So what was the initial y-direction speed? If the acceleration is only in the y-direction, what's the x-direction acceleration?

Hint: zero is a perfectly good value for a parameter

7. Jun 16, 2013

### Robertoalva

so basically

vy = viy +ay t = 0 + (.9)(2.9) and because i have already the velocity of the x direction i can do the v=sqrt( x^2 + y^2) right?

8. Jun 16, 2013

### Staff: Mentor

Yup. That's right.