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How many revolutions will the car go through?

  • #1

Homework Statement


A car starts from rest on a curve with a radius of 190m and accelerates at 0.800 m/s^2 . How many revolutions will the car have gone through when the magnitude of its total acceleration is 2.60 m/s^2 ?


Homework Equations





The Attempt at a Solution



I have tired the magnitude of acceleration = sqrt ((at^2/r^2) +a^2)
so then I said 2.6 =.8t^2/190^2 + .8^2 So I got 297 seconds for the time.
Then I plugged time into x=1/2at^2 but this is where I am not sure. Which acceleration should I be plugging into that equation. I have tried a=2.6 a=.8 and a=a/r =.004 and none of those are giving me a correct answer.

I have been dividing my answer by 2pi to get the answer in revolutions but the stupid online thing keeps saying try again and I am not sure what I am doing wrong. I appreciate any help! Thanks!
 
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Answers and Replies

  • #2
SammyS
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Units! Where are your units?
 
  • #3
Units! Where are your units?
Sorry about that. I just copy pasted so I am not sure why the units didn't show up but I edited it so there should now be units.
 
  • #4
SammyS
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Homework Statement


A car starts from rest on a curve with a radius of 190m and accelerates at 0.800 m/s2 . How many revolutions will the car have gone through when the magnitude of its total acceleration is 2.60 m/s2 ?

Homework Equations


The Attempt at a Solution



I have tired the magnitude of acceleration = sqrt ((at^2/r^2) +a^2)
so then I said 2.6 =.8t^2/190^2 + .8^2 So I got 297 seconds for the time.
Then I plugged time into x=1/2at^2 but this is where I am not sure. Which acceleration should I be plugging into that equation. I have tried a=2.6 a=.8 and a=a/r =.004 and none of those are giving me a correct answer.

I have been dividing my answer by 2pi to get the answer in revolutions but the stupid online thing keeps saying try again and I am not sure what I am doing wrong. I appreciate any help! Thanks!
Square of the magnitude of the acceleration: |a|2 = at2 + ac2, where at is the tangential component of acceleration and ac is the centripetal acceleration (aka, the normal component).

For your problem, at = 0.800 m/s2.

For an object traveling in a circle of radius R and speed v (v being the magnitude of the velocity), ac is given by:

[tex]a_c=\frac{v^2}{R}[/tex].

What must v be for |a|2 to be: (2.60 m/s2)2 ?
 
  • #5
374
1
Centripetal acceleration also equals r*w^2 so you can find w (angular velocity), tangential acceleration=r*alpha so you can find (alpha) angular acceleration. Then use rotational motion equations to find theta (angle of rotation).
 
  • #6
Square of the magnitude of the acceleration: |a|2 = at2 + ac2, where at is the tangential component of acceleration and ac is the centripetal acceleration (aka, the normal component).

For your problem, at = 0.800 m/s2.

For an object traveling in a circle of radius R and speed v (v being the magnitude of the velocity), ac is given by:

[tex]a_c=\frac{v^2}{R}[/tex].

What must v be for |a|2 to be: (2.60 m/s2)2 ?
I used the equation Vf = V0 + at so, v=at and then I plugged that into |a|2 = at2 + ac2 then I solved for t again and this time I got 587 seconds. I don't know if I am getting closer to the answer or further away.
 
  • #7
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1
w^2 (final)=w^2(initial)+2*angular acceleration*theta, w^2(initial)=0 since the car starts at rest. You can calculate angular velocity and angular acceleration (see previous post) so you can find angle of rotation (theta). I get theta to be about .27 Revolutions.
 
  • #8
SammyS
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Homework Statement


A car starts from rest on a curve with a radius of 190m and accelerates at 0.800 m/s^2 . How many revolutions will the car have gone through when the magnitude of its total acceleration is 2.60 m/s^2 ?

The Attempt at a Solution



I have tired the magnitude of acceleration = sqrt ((((at)^2/r)^2) +a^2)
so then I said 2.6^2 =(.8t^2/190)^2 + .8^2 So I got 297 seconds for the time.
I inserted some parentheses and exponents. Is this the computation you really made?
 
  • #9
I inserted some parentheses and exponents. Is this the computation you really made?
You are right! I redid it out and got t=24.24 Is that the answer you got for time?
 
  • #10
SammyS
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I got 27.something seconds.

If t=24.24 then, v=.8×(24.24)≈19.9 m/s

Then, ac = v2/R ≈ 1.98 m/s2.

Need ac=√(a2-at2) ≈ 2.5 m/s2.
 
  • #11
I got 27.something seconds.

If t=24.24 then, v=.8×(24.24)≈19.9 m/s

Then, ac = v2/R ≈ 1.98 m/s2.

Need ac=√(a2-at2) ≈ 2.5 m/s2.
OK I think 27.8 seems to be right. Then would I use x=1/2at2? What value of a should I use? Should I use alpha= a/r =.8/190 =.004? Then have 1/2*.004*27.82 =1.627/2pi = . 2589 revolutions?
 
  • #12
According to this thing the answer is .245 rev. I guess .2589 isn't close enough. IT'S BOGUS
 
  • #13
SammyS
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OK I think 27.8 seems to be right. Then would I use x=1/2at2? What value of a should I use? Should I use alpha= a/r =.8/190 =.004? Then have 1/2*.004*27.82 =1.627/2pi = . 2589 revolutions?
That's a lot of round off: α = a/r =.8/190 =.004 Use 0.00421 and see what you get.

Oh! I see that you actually did use the result without the round-off.

Using the very rounded-off value of .004 rad/s2, I get very close to that BOGUS answer. (0.246 rev.)
 

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