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Anikate

  • one year ago

A driver in a car traveling at a speed of 75 km/h sees a cat 125 m away on the road. How long will it take for the car to accelerate uniformly to a stop in exactly 122 m? Answer in units of s.

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  1. Anikate
    • one year ago
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    any clue??

  2. anonymous
    • one year ago
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    @pooja195 Help me

  3. anonymous
    • one year ago
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    @pooja195 im new to helping

  4. Anikate
    • one year ago
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    do u know the answer??

  5. Anikate
    • one year ago
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    @DylanisNOW u know the answer??

  6. anonymous
    • one year ago
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    Im trying to get @pooja195

  7. Anikate
    • one year ago
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    i thuink she offline, but do know the answer to this question??

  8. Anikate
    • one year ago
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    she appeears to be offline??

  9. Anikate
    • one year ago
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    yea she offline

  10. anonymous
    • one year ago
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    Hold on...Im trying to figure it out. Please dont rush!

  11. Anikate
    • one year ago
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    ok

  12. anonymous
    • one year ago
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    @pooja195 No she isnt

  13. Anikate
    • one year ago
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    ok, do u know how to do it though?

  14. anonymous
    • one year ago
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    Give me a sec please

  15. Anikate
    • one year ago
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    ok np

  16. Anikate
    • one year ago
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    @pgpilot326 do u know how to do this? its a physics question

  17. anonymous
    • one year ago
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    Think of the units.

  18. Anikate
    • one year ago
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    ok now?

  19. anonymous
    • one year ago
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    units are 75km/h right?

  20. Anikate
    • one year ago
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    yes

  21. anonymous
    • one year ago
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    Wait sorry hold on thats not it thats only 1

  22. Anikate
    • one year ago
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    ok

  23. anonymous
    • one year ago
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    75km/h / 125 m=?

  24. Anikate
    • one year ago
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    600m/h

  25. anonymous
    • one year ago
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    Okay do you have answer choices?

  26. Anikate
    • one year ago
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    nopee

  27. Anikate
    • one year ago
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    do u havea roungh estimate of what the andswer could be??

  28. anonymous
    • one year ago
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    Think of it like this. Every second the speed decreases by?

  29. anonymous
    • one year ago
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    We cant give answers...Its just help to get you to understand it!

  30. anonymous
    • one year ago
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    Think of it like this. Every second the speed decreases by?

  31. Anikate
    • one year ago
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    the speed decreases by seconds(squared)

  32. anonymous
    • one year ago
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    Ok that is?

  33. Anikate
    • one year ago
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    idk.....

  34. anonymous
    • one year ago
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    .... @pgpilot326 can you help me help him?

  35. anonymous
    • one year ago
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    Anyways..

  36. Anikate
    • one year ago
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    dylanis, can u just post a step by step explaination for it

  37. anonymous
    • one year ago
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    Ok

  38. anonymous
    • one year ago
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    Step 1:units are 75km/h

  39. anonymous
    • one year ago
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    Step 2:75km/h / 125=600m/h

  40. anonymous
    • one year ago
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    Okay now we are at step 3

  41. anonymous
    • one year ago
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    if a is a constant, integrate to get velocity and again to get position. the velocity should be linear and the position a quadratic.

  42. anonymous
    • one year ago
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    since you know the initial and final speed you should be able to calc. the time

  43. anonymous
    • one year ago
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    pgpilot should we make a equation?

  44. anonymous
    • one year ago
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    should make 2 equations... velocity and position. you'll need both.

  45. Anikate
    • one year ago
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    can you please show bothh?

  46. Anikate
    • one year ago
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    and solve with steps please?

  47. anonymous
    • one year ago
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    \[v \left( t \right)=v_0+at\]and\[x \left( t \right)=x_0+v_0t+\frac{ 1 }{ 2 }at^2\]

  48. anonymous
    • one year ago
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    pg you are right good job btw :)

  49. anonymous
    • one year ago
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    at t=0, initial speed (\(v_0\)) is 75 km/h and initial position is 0 m. at t=\(t_1\), final speed (\(v_f\)) is 0 km/h and final position is 122 m.

  50. anonymous
    • one year ago
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    you need to solve for \(t_1\)

  51. Anikate
    • one year ago
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    ok

  52. Anikate
    • one year ago
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    how so??

  53. Anikate
    • one year ago
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    which equation am i putting all this in??

  54. anonymous
    • one year ago
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    phone, just a minute

  55. anonymous
    • one year ago
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    \[v \left( t_1 \right)=0=75 +a\cdot t_1\] \[x \left( t_1 \right)=122=0+75t_1+\frac{ 1 }{ 2 }at_1^2\]

  56. anonymous
    • one year ago
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    find roots of quadratic in terms of a.

  57. Anikate
    • one year ago
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    u can handwrite it if u wish, ill prob be able to decode it

  58. Anikate
    • one year ago
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    |dw:1443142704493:dw|

  59. Anikate
    • one year ago
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    isnt htat what we have now?? what now??

  60. Anikate
    • one year ago
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    can u show rest of steps

  61. anonymous
    • one year ago
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    solve velocity eqn for t1 and position for t1 and equate

  62. anonymous
    • one year ago
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    to solve for a. once you have a you can find t1

  63. Anikate
    • one year ago
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    how do u solve for t1?? there are 2 variables in the equation itself

  64. anonymous
    • one year ago
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    from velocity: t1 = -75/a

  65. anonymous
    • one year ago
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    use quadratic to solve the position: \[0=-122+75t_1+\frac{ 1 }{ 2 }at_1^2\]for t1 in terms of a

  66. Anikate
    • one year ago
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    |dw:1443143049231:dw|

  67. Anikate
    • one year ago
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    |dw:1443143204081:dw| thats as far as i can get

  68. Anikate
    • one year ago
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    can u please explain how u would do it, and go past that step

  69. anonymous
    • one year ago
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    |dw:1443143218606:dw|

  70. anonymous
    • one year ago
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    use the quadratic formula

  71. anonymous
    • one year ago
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    |dw:1443143430909:dw|then solve for a

  72. Anikate
    • one year ago
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    cant root a variable

  73. Anikate
    • one year ago
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    keep going

  74. anonymous
    • one year ago
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    you'll have to square to get rid of square root

  75. Anikate
    • one year ago
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    ok then, can u plz post work till answer?

  76. Anikate
    • one year ago
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    kinda hard to follow step by step

  77. anonymous
    • one year ago
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    k but it will take a minute

  78. Anikate
    • one year ago
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    take ur time

  79. anonymous
    • one year ago
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    \[122=75\left( \frac{ -75 }{ a } \right)+\frac{ a\left( -75 \right)^2 }{ 2a^2} \Rightarrow 122 =-\frac{ 5625 }{ a } +\frac{ 5625 }{ 2a }\Rightarrow 244a = -11250+5625\]

  80. anonymous
    • one year ago
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    \[a=-\frac{ 5625 }{244 } \Rightarrow t_1 = \frac{ -75 }{ -\frac{ 5625 }{ 244 } }=\frac{ 244 }{ 75} \text{ sec}\]

  81. Anikate
    • one year ago
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    AAH ok! I see! thanks a lot! @pgpilot326

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