Carlos runs with a velocity

[gbedenba]

1. The problem statement, all variables and given/known data
Carlos runs with a velocity of = (6 m/s, 25° north of east) for 10 minutes. How far to the north of his starting position does Carlos end up?

2. Relevant equations



3. The attempt at a solution

I thought maybe you would solve this problem like you do for x and y components? if so...where does the 10 min. go?

[kuruman]

You do have to solve considering x and y components. The time goes in the relevant kinematics equations of motion. What are they?

[gbedenba]

V = Vo + at
X - Xo = Vot + .5at2
v2 = vo2 + 2a(X - Xo)
X - Xo = .5(Vo + V)t

[kuruman]

There is also a similar set of equations for the y direction. What should you use for the acceleration in the x and y direction?

[gbedenba]

i have no clue...i am confused

[kuruman]

Read the problem. Draw North-South and East-West axes. Can you picture in your mind how Carlos is moving? Reread the problem. Is the Carlos' sped increasing, decreasing or staying the same? Does his direction of motion change?

[gbedenba]

I think his speed is staying the same... doesn't his direction stay the same as well

[kuruman]

Yes, his speed and direction stay the same. This means that his velocity does not change. If his velocity does not change, what can you say about his acceleration?

[gbedenba]

acceleration is zero

[kuruman]

Very good. Now put zero in the place of acceleration a in the kinematic equations and write one set for the x-direction and a second set for the y-direction.

[gbedenba]

which equation do I use? is there a way you could show me the first step?

[kuruman]

V = Vo + at becomes V = Vo + 0*t becomes V = Vo.

Now that you know that V = Vo, replace V in the other three equations and replace a with zero like I did above.

X - Xo = Vot + .5at2
v2 = vo2 + 2a(X - Xo)
X - Xo = .5(Vo + V)t

Then repeat with y replacing x everywhere.