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a car is travelling at a constant speed in a circle with radius 100 m
drawing is not quite clear, could you please draw it again? :P
|dw:1378457063394:dw| those are 2 degree
Okay, let's take a look... do we also have a measure for this angle: The angle between r1 and r2?
ya it is 2 degree too... between r1 and r2, r2 and r3, r4 and r5, and r5 and r6 those are all 2 degree
it looks like first we'll need to find the speed of the car going around the circle, by using the equation relating speed, time and distance:
100 m / 0.6
We were given the time to get from r1 to r3... and can also find the distance along the circle between those points, since we were given the circle's radius. But before we can get the distance between the r1 and r3, we'll need the distance around the whole circle (its circumference) Do you remember the formula for circumference?
2 pi r
circumference = 628
its involved a little more than that, for the distance we need to do more calculations.
(628*2)/360 = distance between r1 and r2 = 3.49
then, we see that between each two points marked along the circle, that takes up exactly 1/12th of the whole circumference. Since the points are separated by equal angles...
hmm, are we sure the angles aren't marked as a symbol like: rather than 2 that would make a little more sense based on the picture
sorry but distance between 2 points is 4 degree (r1 to r3) which is 1/90 not 1/12
if it really is 2, then what you have is correct.
ya it is 2
okay, yes if those really are just 2 degrees each I was interpreting it a little differently, but it's difficult to say without seeing the exact problem statement.
thats the exact problem statement i wrote and the exact same figure
If those really are 2 degrees, then the distance between r1 and r3 would be given by 628*4/360, similar to what you wrote.
Now we can plug in this distance and time into the formula speed
yep can u go ahead and solve it please so if i have question i can ask later
So plugging in d= 3.49 and t=.6, we get s=3.49/.6 Could you tell me what that equals?
wait that is for 2 degree, for 4 degree it is 6.9
okay, thanks for checking
np...hey i havee real short time to solve that question like 30 minutes....can you do it asap
Alright, so that gives a speed around the circle 11.5 m/2 at point r2, the car is moving straight up since that's the point where it crosses the axis...
so what are the components
So at that point, the car's horizontal velocity is 0 and the vertical velocity is the full of 11.5 m/s
yep at r2 right
I've circled it on that board
''components'' just means the horizontal and vertical velocities written separately
and yes you're right! :)
Any other questions about this problem?
z componenet at r2?
0... since the car is travelling in a flat circle, it only has x and y components.
nope lol 3 more question if u dont mind
0.6 seconds pass between t_4 when car is at r_4 and t_6 when car is at r_6 determine x,y,z components of car's velocity vector when it is near r_5
Not right now, but maybe later. Good luck with your studies!
I'm tired, sorry. lol :/
thx for explaining stuff
no worries, my pleasure. :P