<li>At time t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is (9.00 m/s^2) i + (4.00 m/s^2) j. It moves at constant speed. At time t2 = 3.00 s (3/4 of a revolution later), its acceleration is (4.00 m/s^2) i + (-9.00 m/s^2) j. What is the radius of the path taken by the particle?</li>
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<p>My approach:
sqrt(9^2 + 4^2) = sqrt(97) = 9.8489.
T=(2pi(r))/v) -----> 3/4 = (2pi(r))/9.8489 -----> v = 1.18 m.
[I think this is wrong. Is my period suppose to be 3/4? Also, how do I get the velocity?]</p>
<li>A boy whirls a stone in a horizontal circle of radius 1.8 m and at height 2.5 m above ground level. The string breaks, and the stone flies horizontally and strikes the ground after traveling a horizontal distance of 10 m. What is the magnitude of the centripetal acceleration of the stone while in circular motion?</li>
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<p>My approach:
y-y0 = v0t + 1/2 *at^2
2.5m = 0 + 1/2 *(-9.8 m/s^2)(t^2)
t = 2.26 s.</p>
<p>y-y0 = v0t + 1/2 *at^2
2.5m = v0(2.26s) + 1/2 * (-9.8 m/s^2)(2.26 s)^2
v0 = 12.2 m/s.</p>
<p>a = v^2/r -----> ((12.2 m/s)^2)/1.8 m = 82.7 m/s^2</p>
<p>[My acceleration for this problem seems a bit large. My approach in the beginning may be incorrect.]</p>
<li>Snow is falling vertically at a constant speed of 8.0 m/s. At what angle from the vertical do the snowflakes appear to be falling as viewed by the driver of a car traveling on a straight road with a speed of 66 km/h?</li>
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<p>[I don’t really understand this problem. I’m not really sure what angle they’re looking for.]</p>
<p>Help me please. Thanks.</p>
<p>for #3 you are overcomplicating things. You found the correct time, but just plug it into the equation dx=vxt to get the tangential velocity (Since the stone will travel in a horizontal direction what a velocity of whatever its tangential velocity was) then plug the vx into v^2/r to get the acceleration</p>
<p>Thanks. That makes a lot of sense now. Any ideas for the other 2 questions?</p>
<p>Wait... for number 3:</p>
<p>y-y0 = v0t + 1/2 *at^2
2.5m = 0 + 1/2 *(-9.8 m/s^2)(t^2)
t = 2.26 s</p>
<p>d=vt -----> v=d/t
10m / 2.26s = 4.425 m/s</p>
<p>a= v^2/r
(4.425 m/s)/1.8 m = 10.9 s. [Answer must be in 3 significant figures.]</p>
<p>When I type 10.9 in as the answer, it says it is incorrect. What did I do wrong here?</p>
<p>I don't think my time is right. I don't even have v0 (initial velocity), so how can I use that equation to find time?</p>
<p>Heck, why not? I guess I could take a stab at the others. I've got a physics test on Thursday and I could use some practice.</p>
<p>For #9, if you were driving a car, and someone dropped something, you wouldn't see a straight vertical path, but rather you would see it as if it were thrown at an angle. Think of it as a large vertical pole that extends from the center of the car. They want you to find the angle between the pole and the path of the snowflake.</p>
<h1>2 Check your units. What you found isn't the velocity. You found the magnitude of the acceleration. The distance traveled seems reasonable- 3/4 a revolution or 3/4 a circle, which you can say is 2(pi)r(3/4)=1.5(pi)r is the distance that was traveled in a time interval of one second, which means the speed is v=1.5(pi)r. Now just substitute into the equation a =v^2/r and now you can find r easily now.</h1>
<p>And by the way, it appears you don't get what a period is. A period is the time it takes the particle to travel one full revolution, so using that equation would get you wrong results. What should be used is the distance travelled, because doing this would get you the velocity.</p>