## 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.

1. Anikate

any clue??

2. anonymous

@pooja195 Help me

3. anonymous

@pooja195 im new to helping

4. Anikate

5. Anikate

6. anonymous

Im trying to get @pooja195

7. Anikate

i thuink she offline, but do know the answer to this question??

8. Anikate

she appeears to be offline??

9. Anikate

yea she offline

10. anonymous

Hold on...Im trying to figure it out. Please dont rush!

11. Anikate

ok

12. anonymous

@pooja195 No she isnt

13. Anikate

ok, do u know how to do it though?

14. anonymous

15. Anikate

ok np

16. Anikate

@pgpilot326 do u know how to do this? its a physics question

17. anonymous

Think of the units.

18. Anikate

ok now?

19. anonymous

units are 75km/h right?

20. Anikate

yes

21. anonymous

Wait sorry hold on thats not it thats only 1

22. Anikate

ok

23. anonymous

75km/h / 125 m=?

24. Anikate

600m/h

25. anonymous

Okay do you have answer choices?

26. Anikate

nopee

27. Anikate

do u havea roungh estimate of what the andswer could be??

28. anonymous

Think of it like this. Every second the speed decreases by?

29. anonymous

We cant give answers...Its just help to get you to understand it!

30. anonymous

Think of it like this. Every second the speed decreases by?

31. Anikate

the speed decreases by seconds(squared)

32. anonymous

Ok that is?

33. Anikate

idk.....

34. anonymous

.... @pgpilot326 can you help me help him?

35. anonymous

Anyways..

36. Anikate

dylanis, can u just post a step by step explaination for it

37. anonymous

Ok

38. anonymous

Step 1:units are 75km/h

39. anonymous

Step 2:75km/h / 125=600m/h

40. anonymous

Okay now we are at step 3

41. anonymous

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

since you know the initial and final speed you should be able to calc. the time

43. anonymous

pgpilot should we make a equation?

44. anonymous

should make 2 equations... velocity and position. you'll need both.

45. Anikate

46. Anikate

47. anonymous

$v \left( t \right)=v_0+at$and$x \left( t \right)=x_0+v_0t+\frac{ 1 }{ 2 }at^2$

48. anonymous

pg you are right good job btw :)

49. anonymous

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

you need to solve for $$t_1$$

51. Anikate

ok

52. Anikate

how so??

53. Anikate

which equation am i putting all this in??

54. anonymous

phone, just a minute

55. anonymous

$v \left( t_1 \right)=0=75 +a\cdot t_1$ $x \left( t_1 \right)=122=0+75t_1+\frac{ 1 }{ 2 }at_1^2$

56. anonymous

find roots of quadratic in terms of a.

57. Anikate

u can handwrite it if u wish, ill prob be able to decode it

58. Anikate

|dw:1443142704493:dw|

59. Anikate

isnt htat what we have now?? what now??

60. Anikate

can u show rest of steps

61. anonymous

solve velocity eqn for t1 and position for t1 and equate

62. anonymous

to solve for a. once you have a you can find t1

63. Anikate

how do u solve for t1?? there are 2 variables in the equation itself

64. anonymous

from velocity: t1 = -75/a

65. anonymous

use quadratic to solve the position: $0=-122+75t_1+\frac{ 1 }{ 2 }at_1^2$for t1 in terms of a

66. Anikate

|dw:1443143049231:dw|

67. Anikate

|dw:1443143204081:dw| thats as far as i can get

68. Anikate

can u please explain how u would do it, and go past that step

69. anonymous

|dw:1443143218606:dw|

70. anonymous

71. anonymous

|dw:1443143430909:dw|then solve for a

72. Anikate

cant root a variable

73. Anikate

keep going

74. anonymous

you'll have to square to get rid of square root

75. Anikate

ok then, can u plz post work till answer?

76. Anikate

kinda hard to follow step by step

77. anonymous

k but it will take a minute

78. Anikate

take ur time

79. anonymous

$122=75\left( \frac{ -75 }{ a } \right)+\frac{ a\left( -75 \right)^2 }{ 2a^2} \Rightarrow 122 =-\frac{ 5625 }{ a } +\frac{ 5625 }{ 2a }\Rightarrow 244a = -11250+5625$

80. anonymous

$a=-\frac{ 5625 }{244 } \Rightarrow t_1 = \frac{ -75 }{ -\frac{ 5625 }{ 244 } }=\frac{ 244 }{ 75} \text{ sec}$

81. Anikate

AAH ok! I see! thanks a lot! @pgpilot326