Help with 2D Motion: Acceleration & Coordinates

[motionman04]

2. [PSE6 4.P.005.] At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of vi = (3.00 i - 2.00 j) m/s and is at the origin. At t = 2.80 s, the particle's velocity is v = (9.10 i + 8.50 j) m/s.
(a) Find the acceleration of the particle at any time t. (Use t, i, and j as necessary.) Got this one


(b) Find its coordinates at any time t.
x = ( ) m
y = ( ) m

Don't get these two.

 

[Pyrrhus]

The problem states constant acceleration, so use the kinematics equations for uniform acceleration, with both the speeds (x-axis and y-axis) and the acceleration(x-axis and y-axis) you found to find the coordinate on the x-axis, and then on the y-axis.

 

[motionman04]

One other question:

To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 330 m/s at 46.0° above the horizontal. It explodes on the mountainside 36.0 s after firing. What are the x and y coordinates of the shell where it explodes, relative to its firing point?

I got the x coordinates by doing 330 cos 46 * 36, but when I try yf = 330sin46 - 1/2(-9.8)(36)^2, the answer is wrong. How do I find the y coordinates?

 

[Pyrrhus]

One other question:
I try yf = 330sin46 - 1/2(-9.8)(36)^2, the answer is wrong. How do I find the y coordinates?

You forgot time on your initial speed


Y - Y_{o} = V_{o}t + \frac{1}{2}at^2

 

[motionman04]

Well, I tried yf = 330sin46(36) - 1/2(-9.8)(36)^2 and it turned out to be about 14896 feet, which was wrong lol, any ideas?

 

[Pyrrhus]

Well, I tried yf = 330sin46(36) - 1/2(-9.8)(36)^2 and it turned out to be about 14896 feet, which was wrong lol, any ideas?

You already has a negative sign for acceleration, it seems to me you're plugging it twice.

 

[motionman04]

oo okay, I just tried yf = 330sin46 + 1/2(-9.8)(36)^2, and got 2195 feet, that sound right?

 

[Pyrrhus]

oo okay, I just tried yf = 330sin46 + 1/2(-9.8)(36)^2, and got 2195 feet, that sound right?

feet? you mean meters, It's the answer yes.

 

[motionman04]

lol thanks, but yeah, one other one that was giving me problems: PSE6 4.P.014.] An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 10.0 m if her initial speed is 2.40 m/s. What is the free-fall acceleration on the planet?

I plugged stuff into the kinematics formula, but not getting it

 

[motionman04]

O never mind

 

[motionman04]

6. [PSE6 4.P.029.] A tire 0.300 m in radius rotates at a constant rate of 190 rev/min. Find the speed of a small stone lodged in the tread of the tire (on its outer edge). (Hint: In one revolution, the stone travels a distance equal to the circumference of its path, 2r.)
m/s
What is the acceleration of the stone?

lol yeah, need help with this one

 

[Pyrrhus]

6. [PSE6 4.P.029.] A tire 0.300 m in radius rotates at a constant rate of 190 rev/min. Find the speed of a small stone lodged in the tread of the tire (on its outer edge). (Hint: In one revolution, the stone travels a distance equal to the circumference of its path, 2r.)
m/s
What is the acceleration of the stone?

lol yeah, need help with this one

Think about the hint, and you will get it.

 

[motionman04]

Well I tried 2pi(.300), but I'm wondering if you have to convert the rev/minute

 

[Pyrrhus]

Well I tried 2pi(.300), but I'm wondering if you have to convert the rev/minute

Let me phrase it this way

1 revolution is 2pir meters.

and you got 190 rev/min, so how many meters will be 190 revs?

 

[motionman04]

2pi(190) i get 1193.8 meters

 

[Pyrrhus]

2pi(190) i get 1193.8 meters

maybe you didn't see the r for radius

1 rev is 2piR meters

190 rev is ?

 

[motionman04]

oo okay, I solved 190=2pir and got 30.239 m

 

[motionman04]

I divided that by 60 to get .504 m/s, where do I go from here?