Particle with horizontal velocity and vertical acceleration

[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?

Homework Equations

v= vi +at

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.

 

[gneill]

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.

 

[Robertoalva]

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

 

[UVW]

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

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

 

[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?

 

[gneill]

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

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

how do i get the missing values?

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

 

[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?

 

[gneill]

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?

Yup. That's right.