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KnitWit Group Title

I'm a homeschooler, using BJU Press's curriculum. It's normally great, but with my physics I'm struggling and can't seem to find any help. I'm in the kinematics section, dealing with motion in a plane and projectile motion and such, and I don't understand much of it. I need a lot of help with this. Please, if anyone could. This would be a sample question (I think). If a car is traveling south at a velocity of 110 km/hr, and after 2 minutes, is traveling west at a velocity of 110 km/hr, what was its acceleration throughout the curve? There are also other kinds of questions in the same category which I totally don't get either, but this is an ok one to just start out with. Thank you for reading this.

  • one year ago
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  1. BadgalRiRi Group Title
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    Hi Im a homeschooler too. Soo true totally struggling

    • one year ago
  2. KnitWit Group Title
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    :)

    • one year ago
  3. ivancsc1996 Group Title
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    Okay this is hard. Have you seen uniform circular motion? Well basically and acceleration is a vector that changes velocity. But remember that velocity is a vector as well so it can be changed in two ways (the two components of a vector): magnitude and direction. So in this case there is a centripetal acceleration that changes the direction of the velocity but not its magnitude. Remember that to have an acceleration you need a force that causes it, in this case it is the friction of the wheels with the paviment, (some extra knowledge). The centripetal Force can be calculated by multiplying the mass with the velocity squared and the dividing by the radius. Then you know that force is mass times acceleration so if you divide everything by the mass you get an equation for acceleration You would get this equation:\[F _{cent}=\frac{ mv ^{2} }{ r }=ma _{cent} \rightarrow a _{cent}=\frac{ v ^{2} }{ r }\]You can calculate the radius with some geometry. If the car starts looking south and then ends up looking west it made a quarter of a circle turn. You can calculate that distance which is a fourth of the total circunference by multiplying the velocity times the time it took. The you know that the circunference is two times pi times r. If you do some algebra you get this equation for r:\[\frac{ C }{ 4 }=vt \rightarrow C=4vt; C=2\pi r \rightarrow4vt=2\pi r \rightarrow r=\frac{ 2vt }{ \pi }\]Then if you substitute that to our previous equation you get this answer:\[a _{cent}=\frac{ v ^{2} }{ \frac{ 2vt }{ \pi }}=\frac{ v \pi }{ 2t }=\frac{ 110\frac{ km }{ h }\frac{ 1000m }{ 1km } \frac{ 1h }{ 3600s }\pi}{ 2\min \frac{ 60s }{ 1\min }2 }=0.3999712869\frac{ m }{ s ^{2} }=0.4\frac{ m }{ s ^{2} }\]If you haven't seen uniform circular motion maybe what the are asking is the tangential acceleration which is the one that changes the magnitude of the velocity but since there is no magnitude change then it must be zero. So your answer is 0.\[a _{\tan}=0\]|dw:1366665592363:dw|

    • one year ago
  4. KnitWit Group Title
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    This is very, very confusing.

    • one year ago
  5. ivanmlerner Group Title
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    It doesnt say its a circular motion, therefore the aceleration that he is asking for is the average. The aceleration tells us how the speed change with time, so it is:\[\frac{\Delta V}{\Delta t}=a\]Now we need to look at the begining and the end of the motion. Since aceleration is a vector, we can separate it into components. On the south component we see that there was a velocity of 110km/h and now it is 0 (since its not moving south anymore, the velocity in this direction is 0). The aceleration that we get on the south component is: \[\frac{\Delta V}{\Delta t}= \frac{110}{1/30}=3300km/h^2\]The 1/30 instead of 2, is because we need to use the same units, and the velocity is in km/h so the time must be in hr. Doing the same thing with the west component we get to this graph with the two components, and summing them we get the final acceleration:|dw:1366903445705:dw|Looking at the drawing, we see that the final aceleration is given by: \[a=\sqrt{3300^2+3300^2}=3300\sqrt{2}km/h^2\]To convert to the standard units, we do this:\[\frac{km}{h^2}=\frac{1000m}{(60^2s)^2}=\frac{1000m}{12960000s^2}=\frac{1}{12960}\frac{m}{s^2}\]So, the aceleration is:\[a=3300\sqrt{2}km/h^2=\frac{3300}{12960}\sqrt{2}m/s^2=\frac{55}{216}\sqrt{2}m/s^2=0.36m/s^2\]That is the answer, but in a more understandable unit it is:\[a=1,3\frac{km/h}{s}\]This tells us that at each second that passes, the speed changes 1.3 km/h, wich is possible for a car. Just remember that this change might not be in the velocity shown in the velocimeter, because it is a vector and the aceleration can only change its direction.

    • one year ago
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