Vertical acceleration and horizontal velocity

[Robertoalva]

1. (a) A particle travels horizontally between two parallel walls separated by 18.4 m. It moves toward the opposing wall at a constant rate of 9.4 m/s. Also, it has an acceleration in the direction parallel to the walls of 3.6 m/s^2, what will be it's speed when it touches the wall?

(b)At what angle with the wall will the particle
strike?

Homework Equations

v^2= vi^2 +2a(x-xi)

The Attempt at a Solution

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(a) the thing is like this, the body diagram shows that the particle takes a curve that is similar to a parabola, and that messes my mind.

I think they are asking just for the final velocity. correct me if I'm wrong, please.

 

[darkxponent]

I think they are asking just for the final velocity. correct me if I'm wrong, please.

Yes they are asking for the final velocity. Show us some attempt so that we can help you with that.

 

[barryj]

The question states.."(b)At what angle with the wall will the particle
strike? ' What is the angle?

 

[Robertoalva]

v^2=v^2+2a(x-xi)
v^2=(9.4)^2 +2(3.6)(18.4)
v^2=(88.33)+(132.48)
v^2=220.81
v=sqrt(220.81)

v=14.85 m/s

There's my final velocity, now to get the angle, do I use a formula for a projectile?

 

[barryj]

I don't think your vertical velocity is correct.
Ask yourself, how long to go from one wall to the other, then v = at.

Now with the two velocities, you can easily find the angle

 

[Robertoalva]

Well, if I have to find a vertical velocity, I don't have time. And from one wall to another it says that the separation between the walls is 18.4 m.

 

[HallsofIvy]

v^2=v^2+2a(x-xi)
v^2=(9.4)^2 +2(3.6)(18.4)
v^2=(88.33)+(132.48)
v^2=220.81
v=sqrt(220.81)

v=14.85 m/s

There's my final velocity, now to get the angle, do I use a formula for a projectile?

You are confusing a number of things. First, technically, velocity is a vector not a number. And you are combining "horizontal" and "vertical". The "18.4 m" distance is horizontal while the "4.6 m/s^2" acceleration if vertical.

Use the 18.4 m distance and 9.4 m/s constant horizontal speed to determine the time the particle takes . Use that time with the 4.6 m/s^2 vertical acceleration to determine the vertical speed when the particle hits the wall. Once you have both horizontal and vertical speeds when the particle hits the wall, use the fact that the tangent of the angle is "horizontal speed divided by vertical speed".

 

[Robertoalva]

so for the horizontal part, I use x=xi + vit +1/2at^2 ? and vi=9.4 and v=9.4 also?

 

[barryj]

No, first find the time it takes to go from one wall to the other in the horizontal direction. Then once you have the time of travel, you can find the vertical velocity at the opposite wall using v = at.

 

[Robertoalva]

please disregard the last post of mine, it was a brain fart. lol

 

[Robertoalva]

i got t= 3.19s whit the formula sqrt(2(x-xi)/a) = t am I wrong?

 

[barryj]

You are complicating the problem. To get the horizontal time, use t = d/v. What is the distance between the walls? what is the horizontal velocity?

Then when you have the horizontal time use this time to find the vertical velocity. v = at.

 

[Robertoalva]

ok, thanks. and wouldn't it be v= vi+ at?

 

[barryj]

The initial vertical velocity is 0 so its just v = at

 

[Robertoalva]

i tried with v=vi+at and with v= at and it gives me the wrong answer what I'm doing wrong!?

 

[barryj]

What did you get for the time?
Then what did you get for the vertical velocity?

But remember, the answe3r is an angle not velocity.

 

[Robertoalva]

in the part (a) they are asking me to find the speed of the particle when it touches the other wall.

t= d/v= (18.4m)/(9.4m/s)= 1.95s

v= at = (3.6m/s^2)(1.95s)= 7.02 m/s

 

[barryj]

Correct. Now you have the horizontal velocity, 9.4 m/s and the vertical velocity of 7.02 m/s.
HOw do you find the angle?

 

[Robertoalva]

but duuuuuude! I'm telling you, they are asking me first for the speed when it touches the other wall! i need that first do i add both velocities?

 

[barryj]

You are correct. Speed and angle. You have two vectors that are 90 degrees apart. 9.4 in the x direction and 7.02 in the y direction. What is the resultant speed? Use the Greek formula. Then get the angle

 

[Robertoalva]

so the speed would be just adding both velocities and for the angle is tan= y/x right?

 

[barryj]

the angle is correct but add the horizontal and vertical velocities vectorially.
Be careful which angle is requested. They are asking for angle with the wall!

 

[Robertoalva]

ok i got the velocity, and the angle is with the wall so it means y-axis right? and it would be sin^-1(7.02) ?

 

[barryj]

The angle with the horizontal will be tan^-1(y/x) but the angle with the wall will be 90 - the angle with the horizontal.

 

[Robertoalva]

thanks ! this problem is finally over!