How Fast Does a Penguin Slide on Ice After 8.75 Seconds?

[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?

 

[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?

 

[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.

 

[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.

 

[jambaugh]

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

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