Homework Help: Penguin: Kinematics Problem

1. Jan 27, 2008

anastasiaw

A car that weighs 12000.0 N is initially moving at a speed of 30.0 km/hr when the brakes are applied and the car is brought to a stop in 4.1 s. Find the magnitude of the force that stops the car, assuming it is constant.
This is the wrong problem ^^^

A 6.70 kg penguin runs onto a huge horizontal sheet of frictionless ice. The sheet lies on the xy-plane. At t = 0 s it is at x = 0 m and y = 0 m with an initial speed of 0.75 m/s along the positive x-axis. It slides while being pushed by the wind with a force of 0.37 N directed along the positive y-axis. Calculate the penguin's speed at t = 8.75 s.

F=ma => 0.37=6.7a => a=0.6 m/s^2
vy(t )= vy0 + at = 0 + .06(8.75) = 0.483 m/s
v(t)^2 = vx(t)^2 + vy(t)^2 = 0.75^2 + 0.483^2 = 0.796
v(t) = 0.892 m/s
This part is CORRECT.

Calculate the direction of his velocity at that time. Give the angle with respect to the x-axis.

Set the vectors up as a right triangle, find the angle between the resultant vector and the x-vector.

x-vector: 0.75 m/s
y-vector: 0.483 m/s
angle should be: sin^-1(y/x) = sin^-1(.483/.75) = 40.1 deg
This part is INCORRECT.

What did I do wrong?

Last edited: Jan 27, 2008
2. Jan 27, 2008

jambaugh

I don't see why you're setting up right angles. The problem appears one dimensional to me. Are you leaving something out? Did you mean acceleration instead of velocity?

3. Jan 27, 2008

anastasiaw

"Draw the components of the resultant velocity and extract the answer from the right triangle you obtain. The angle must be between 0 and 2 pi radians."

That's what it told me when I entered the incorrect answer. I'm taking the x-component of the velocity and the y-component and setting them up tip-to-tail. The resultant vector should be the magnitude and direction of the velocity I was looking for in the first part (and I got the magnitude part right).

Edit: I'm sorry -- I wrote the first part of the problem wrong. Actually I copied the wrong problem. I'll fix that.

4. Jan 28, 2008

anastasiaw

Figured it out: for some reason I was thinking "soa" instead of "toa" -- tan of an angle = opposite/adjacent

So I should have done arctan (y/x) which yields 32.8 deg; this is the correct answer.

5. Jan 28, 2008

jambaugh

Ah yes, I suspected something off. Glad you got it worked out on your own.