Penguin Sliding on Antarctic Ice: Velocity at t=11.81s

[PierceJ]

Homework Statement

A 5.1 kg penguin runs onto a huge sheet of frictionless Antarctic ice. At t=0 it is at x=0 and y=0 with an initial velocity of 0.39 m/s along the positive x-axis. It slides while being pushed by the wind with a force of 0.53 N directed along the positive y-axis. Calculate the magnitude of the penguin's velocity at t = 11.81 s.

Homework Equations

x = V_0xt
y = V_oyt + 1/2a_yt²

The Attempt at a Solution

The penguin is traveling in the positive x direction and being pushed up in the positive y direction, so I need to find the horizontal distance and the vertical distance and then I can get the magnitude and angle from that. So I start by setting up a drawing on a coordinate plane where the penguin starts at the origin and it goes 0.39(11.81)m in the x direction but I don't know what I should do for the y direction. Any help is greatly appreciated.

 

[Chestermiller]

The question is asking for the velocity, not the distance. What is the acceleration in the y direction? If this acceleration continues for 11.81 sec, what is the velocity in the y direction after 11.81 sec. What is the resultant velocity?

Chet

 

[PierceJ]

So the acceleration in the y direction is:
F = ma
0.53 = 5.1a
a = 0.1039m/s

So then I use V = V₀ + at?
But what is the initial velocity in the y direction?

 

[Chestermiller]

What do the words "with an initial velocity of 0.39 m/s along the positive x-axis" mean to you?

Chet

 

[PierceJ]

It means that its moving 0.39 m/s horizontally.

 

[Chestermiller]

It means that its moving 0.39 m/s horizontally.

It means that its initial velocity in the y direction is zero.

Chet

 

[PierceJ]

Oh, I see now.
V = V₀ + at
V = 1.288m/s

So then for the second part of the problem, I need to calculate the angle of that velocity.
So I have to find the displacement in x and then do the inverse cosine for the angle?

 

[Chestermiller]

Oh, I see now.
V = V₀ + at
V = 1.288m/s

So then for the second part of the problem, I need to calculate the angle of that velocity.
So I have to find the displacement in x and then do the inverse cosine for the angle?

The direction of the displacement is not the same as the direction of the velocity. So why are you asking about the displacement? Do you know how to determine the magnitude of a vector if you know its two perpendicular components?

Chet

 

[PierceJ]

Isnt it root(x²+y²)?

I'm asking because the question is asking me to find the angle.

 

[Chestermiller]

Isnt it root(x²+y²)?

I'm asking because the question is asking me to find the angle.

The magnitude of the velocity is $\sqrt{(v_x)^2+(v_y)^2}$

The direction of the velocity is $\tan θ=\frac{v_y}{v_x}$

Note that there are no distances in these equations.

Chet

 

[PierceJ]

So root((.39^2)+(1.288^2)) = 1.346

tan^-1(1.346) = 53.38deg

 

[Chestermiller]

So root((.39^2)+(1.288^2)) = 1.346

tan^-1(1.346) = 53.38deg

The 1.346 is correct. The determination of the angle is not. See the equation for the angle again.

Chet

 

[PierceJ]

Oh oops, read that wrong. Got it now, thanks a bunch!