Angle that car is at after angular acceleration












2












$begingroup$


A car starts from rest on a curve with a radius of $150m$ and tangential acceleration of $displaystyle 1.5frac{m}{s^2}$.



Through what angle will the car have traveled when the magnitude of its total acceleration is $displaystyle 2.2frac{m}{s^2}$?



A friend and I have been working on this for an hour and just cannot get the right answer.



We could use some help finding an equation that might be useful.



We used the pythagorean theorem to get the radial acceleration ($displaystyle 1.6frac{m}{s^2}$) and then used $displaystyle A = frac{v^2}{r}$ to get velocity (we got $displaystyle 15.53 frac{m}{s}$).



We don't know what to do past there.






calculus trigonometry physics mathematical-physics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    There is info missing on how the car gains acceleration.
    $endgroup$
    – CAGT
    Feb 1 '14 at 4:59










  • $begingroup$
    That's all they gave us..
    $endgroup$
    – user38696
    Feb 1 '14 at 5:08










  • $begingroup$
    If the car was hanging on a cliff or vertical loop, we could assume gravity as an acceleration, but since it states it is at rest on a curve, the only other assumption possible is that the question is: What angle will the car have traveled when magnitude of its total velocity is 2.2m/s
    $endgroup$
    – CAGT
    Feb 1 '14 at 5:10












  • $begingroup$
    I'm pretty sure that's what the question is
    $endgroup$
    – user38696
    Feb 1 '14 at 5:18










  • $begingroup$
    I can't help you. Perhaps someone else might understand better what is going on. On my eyes, you need info on how acceleration changes.
    $endgroup$
    – CAGT
    Feb 1 '14 at 5:26
















2












$begingroup$


A car starts from rest on a curve with a radius of $150m$ and tangential acceleration of $displaystyle 1.5frac{m}{s^2}$.



Through what angle will the car have traveled when the magnitude of its total acceleration is $displaystyle 2.2frac{m}{s^2}$?



A friend and I have been working on this for an hour and just cannot get the right answer.



We could use some help finding an equation that might be useful.



We used the pythagorean theorem to get the radial acceleration ($displaystyle 1.6frac{m}{s^2}$) and then used $displaystyle A = frac{v^2}{r}$ to get velocity (we got $displaystyle 15.53 frac{m}{s}$).



We don't know what to do past there.






calculus trigonometry physics mathematical-physics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    There is info missing on how the car gains acceleration.
    $endgroup$
    – CAGT
    Feb 1 '14 at 4:59










  • $begingroup$
    That's all they gave us..
    $endgroup$
    – user38696
    Feb 1 '14 at 5:08










  • $begingroup$
    If the car was hanging on a cliff or vertical loop, we could assume gravity as an acceleration, but since it states it is at rest on a curve, the only other assumption possible is that the question is: What angle will the car have traveled when magnitude of its total velocity is 2.2m/s
    $endgroup$
    – CAGT
    Feb 1 '14 at 5:10












  • $begingroup$
    I'm pretty sure that's what the question is
    $endgroup$
    – user38696
    Feb 1 '14 at 5:18










  • $begingroup$
    I can't help you. Perhaps someone else might understand better what is going on. On my eyes, you need info on how acceleration changes.
    $endgroup$
    – CAGT
    Feb 1 '14 at 5:26














2












2








2





$begingroup$


A car starts from rest on a curve with a radius of $150m$ and tangential acceleration of $displaystyle 1.5frac{m}{s^2}$.



Through what angle will the car have traveled when the magnitude of its total acceleration is $displaystyle 2.2frac{m}{s^2}$?



A friend and I have been working on this for an hour and just cannot get the right answer.



We could use some help finding an equation that might be useful.



We used the pythagorean theorem to get the radial acceleration ($displaystyle 1.6frac{m}{s^2}$) and then used $displaystyle A = frac{v^2}{r}$ to get velocity (we got $displaystyle 15.53 frac{m}{s}$).



We don't know what to do past there.






calculus trigonometry physics mathematical-physics







share|cite|improve this question











$endgroup$




A car starts from rest on a curve with a radius of $150m$ and tangential acceleration of $displaystyle 1.5frac{m}{s^2}$.



Through what angle will the car have traveled when the magnitude of its total acceleration is $displaystyle 2.2frac{m}{s^2}$?



A friend and I have been working on this for an hour and just cannot get the right answer.



We could use some help finding an equation that might be useful.



We used the pythagorean theorem to get the radial acceleration ($displaystyle 1.6frac{m}{s^2}$) and then used $displaystyle A = frac{v^2}{r}$ to get velocity (we got $displaystyle 15.53 frac{m}{s}$).



We don't know what to do past there.






calculus trigonometry physics mathematical-physics




calculus trigonometry physics mathematical-physics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 1 '14 at 4:59









2012ssohn

3,59711029




3,59711029










asked Feb 1 '14 at 4:54









user38696user38696

112




112








  • 1




    $begingroup$
    There is info missing on how the car gains acceleration.
    $endgroup$
    – CAGT
    Feb 1 '14 at 4:59










  • $begingroup$
    That's all they gave us..
    $endgroup$
    – user38696
    Feb 1 '14 at 5:08










  • $begingroup$
    If the car was hanging on a cliff or vertical loop, we could assume gravity as an acceleration, but since it states it is at rest on a curve, the only other assumption possible is that the question is: What angle will the car have traveled when magnitude of its total velocity is 2.2m/s
    $endgroup$
    – CAGT
    Feb 1 '14 at 5:10












  • $begingroup$
    I'm pretty sure that's what the question is
    $endgroup$
    – user38696
    Feb 1 '14 at 5:18










  • $begingroup$
    I can't help you. Perhaps someone else might understand better what is going on. On my eyes, you need info on how acceleration changes.
    $endgroup$
    – CAGT
    Feb 1 '14 at 5:26














  • 1




    $begingroup$
    There is info missing on how the car gains acceleration.
    $endgroup$
    – CAGT
    Feb 1 '14 at 4:59










  • $begingroup$
    That's all they gave us..
    $endgroup$
    – user38696
    Feb 1 '14 at 5:08










  • $begingroup$
    If the car was hanging on a cliff or vertical loop, we could assume gravity as an acceleration, but since it states it is at rest on a curve, the only other assumption possible is that the question is: What angle will the car have traveled when magnitude of its total velocity is 2.2m/s
    $endgroup$
    – CAGT
    Feb 1 '14 at 5:10












  • $begingroup$
    I'm pretty sure that's what the question is
    $endgroup$
    – user38696
    Feb 1 '14 at 5:18










  • $begingroup$
    I can't help you. Perhaps someone else might understand better what is going on. On my eyes, you need info on how acceleration changes.
    $endgroup$
    – CAGT
    Feb 1 '14 at 5:26








1




1




$begingroup$
There is info missing on how the car gains acceleration.
$endgroup$
– CAGT
Feb 1 '14 at 4:59




$begingroup$
There is info missing on how the car gains acceleration.
$endgroup$
– CAGT
Feb 1 '14 at 4:59












$begingroup$
That's all they gave us..
$endgroup$
– user38696
Feb 1 '14 at 5:08




$begingroup$
That's all they gave us..
$endgroup$
– user38696
Feb 1 '14 at 5:08












$begingroup$
If the car was hanging on a cliff or vertical loop, we could assume gravity as an acceleration, but since it states it is at rest on a curve, the only other assumption possible is that the question is: What angle will the car have traveled when magnitude of its total velocity is 2.2m/s
$endgroup$
– CAGT
Feb 1 '14 at 5:10






$begingroup$
If the car was hanging on a cliff or vertical loop, we could assume gravity as an acceleration, but since it states it is at rest on a curve, the only other assumption possible is that the question is: What angle will the car have traveled when magnitude of its total velocity is 2.2m/s
$endgroup$
– CAGT
Feb 1 '14 at 5:10














$begingroup$
I'm pretty sure that's what the question is
$endgroup$
– user38696
Feb 1 '14 at 5:18




$begingroup$
I'm pretty sure that's what the question is
$endgroup$
– user38696
Feb 1 '14 at 5:18












$begingroup$
I can't help you. Perhaps someone else might understand better what is going on. On my eyes, you need info on how acceleration changes.
$endgroup$
– CAGT
Feb 1 '14 at 5:26




$begingroup$
I can't help you. Perhaps someone else might understand better what is going on. On my eyes, you need info on how acceleration changes.
$endgroup$
– CAGT
Feb 1 '14 at 5:26










1 Answer
1






active

oldest

votes


















1












$begingroup$

You are simply trying to get the distance traveled around the curve in order to get the velocity. I think you are correct that radial acceleration is



$$a_r = sqrt{a^2-a_t^2} = sqrt{(2.2)^2-(1.5)^2} frac{text{m}}{text{sec}^2} approx 1.6 frac{text{m}}{text{sec}^2} $$



and you know that



$$a_r = frac{v^2}{r} implies v = sqrt{a_r r} approx 15.5 frac{text{m}}{text{sec}}$$



This speed corresponds to tangential velocity. The distance traveled around the curve is



$$d = frac12 a_t t^2$$



where the time $t$ is known from the relation $v = a_t t$, which means that



$$d = frac{v^2}{2 a_t} = frac{a_r r}{2 a_t}$$



The angle through which the car has traveled is given by $theta = d/r$, or



$$theta = frac{a_r}{2 a_t} = frac12 sqrt{frac{a^2}{a_t^2}-1} approx 0.536 , text{rad} approx 30.7^{circ}$$






share|cite|improve this answer









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    $begingroup$

    You are simply trying to get the distance traveled around the curve in order to get the velocity. I think you are correct that radial acceleration is



    $$a_r = sqrt{a^2-a_t^2} = sqrt{(2.2)^2-(1.5)^2} frac{text{m}}{text{sec}^2} approx 1.6 frac{text{m}}{text{sec}^2} $$



    and you know that



    $$a_r = frac{v^2}{r} implies v = sqrt{a_r r} approx 15.5 frac{text{m}}{text{sec}}$$



    This speed corresponds to tangential velocity. The distance traveled around the curve is



    $$d = frac12 a_t t^2$$



    where the time $t$ is known from the relation $v = a_t t$, which means that



    $$d = frac{v^2}{2 a_t} = frac{a_r r}{2 a_t}$$



    The angle through which the car has traveled is given by $theta = d/r$, or



    $$theta = frac{a_r}{2 a_t} = frac12 sqrt{frac{a^2}{a_t^2}-1} approx 0.536 , text{rad} approx 30.7^{circ}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You are simply trying to get the distance traveled around the curve in order to get the velocity. I think you are correct that radial acceleration is



      $$a_r = sqrt{a^2-a_t^2} = sqrt{(2.2)^2-(1.5)^2} frac{text{m}}{text{sec}^2} approx 1.6 frac{text{m}}{text{sec}^2} $$



      and you know that



      $$a_r = frac{v^2}{r} implies v = sqrt{a_r r} approx 15.5 frac{text{m}}{text{sec}}$$



      This speed corresponds to tangential velocity. The distance traveled around the curve is



      $$d = frac12 a_t t^2$$



      where the time $t$ is known from the relation $v = a_t t$, which means that



      $$d = frac{v^2}{2 a_t} = frac{a_r r}{2 a_t}$$



      The angle through which the car has traveled is given by $theta = d/r$, or



      $$theta = frac{a_r}{2 a_t} = frac12 sqrt{frac{a^2}{a_t^2}-1} approx 0.536 , text{rad} approx 30.7^{circ}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You are simply trying to get the distance traveled around the curve in order to get the velocity. I think you are correct that radial acceleration is



        $$a_r = sqrt{a^2-a_t^2} = sqrt{(2.2)^2-(1.5)^2} frac{text{m}}{text{sec}^2} approx 1.6 frac{text{m}}{text{sec}^2} $$



        and you know that



        $$a_r = frac{v^2}{r} implies v = sqrt{a_r r} approx 15.5 frac{text{m}}{text{sec}}$$



        This speed corresponds to tangential velocity. The distance traveled around the curve is



        $$d = frac12 a_t t^2$$



        where the time $t$ is known from the relation $v = a_t t$, which means that



        $$d = frac{v^2}{2 a_t} = frac{a_r r}{2 a_t}$$



        The angle through which the car has traveled is given by $theta = d/r$, or



        $$theta = frac{a_r}{2 a_t} = frac12 sqrt{frac{a^2}{a_t^2}-1} approx 0.536 , text{rad} approx 30.7^{circ}$$






        share|cite|improve this answer









        $endgroup$



        You are simply trying to get the distance traveled around the curve in order to get the velocity. I think you are correct that radial acceleration is



        $$a_r = sqrt{a^2-a_t^2} = sqrt{(2.2)^2-(1.5)^2} frac{text{m}}{text{sec}^2} approx 1.6 frac{text{m}}{text{sec}^2} $$



        and you know that



        $$a_r = frac{v^2}{r} implies v = sqrt{a_r r} approx 15.5 frac{text{m}}{text{sec}}$$



        This speed corresponds to tangential velocity. The distance traveled around the curve is



        $$d = frac12 a_t t^2$$



        where the time $t$ is known from the relation $v = a_t t$, which means that



        $$d = frac{v^2}{2 a_t} = frac{a_r r}{2 a_t}$$



        The angle through which the car has traveled is given by $theta = d/r$, or



        $$theta = frac{a_r}{2 a_t} = frac12 sqrt{frac{a^2}{a_t^2}-1} approx 0.536 , text{rad} approx 30.7^{circ}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 1 '14 at 6:49









        Ron GordonRon Gordon

        122k14155265




        122k14155265






























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