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- Thread starter Shah 72
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- Mar 1, 2012

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$T = \dfrac{\sqrt{1105}+5}{9} \approx 4.25$

in the attached graph of velocity vs time ...

shaded area above the x-axis = displacement from A to B

shaded area below the x-axis = displacement from B to A

the two shaded areas sum to zero, indicating a return to the starting point, A

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- #3

Thanks so much!$\displaystyle \dfrac{1}{12}\int_0^T 3t^3-16t^2-10t+60 \, dt = 0$

$T = \dfrac{\sqrt{1105}+5}{9} \approx 4.25$

in the attached graph of velocity vs time ...

shaded area above the x-axis = displacement from A to B

shaded area below the x-axis = displacement from B to A

the two shaded areas sum to zero, indicating a return to the starting point, A

View attachment 11188

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- #4

Can I pls ask you$\displaystyle \dfrac{1}{12}\int_0^T 3t^3-16t^2-10t+60 \, dt = 0$

$T = \dfrac{\sqrt{1105}+5}{9} \approx 4.25$

in the attached graph of velocity vs time ...

shaded area above the x-axis = displacement from A to B

shaded area below the x-axis = displacement from B to A

the two shaded areas sum to zero, indicating a return to the starting point, A

View attachment 11188

Can you pls tell me how did you calculate that? Did you solve the cubic equation? I don't understand how you got the equation sq root (1105).$\displaystyle \dfrac{1}{12}\int_0^T 3t^3-16t^2-10t+60 \, dt = 0$

$T = \dfrac{\sqrt{1105}+5}{9} \approx 4.25$

in the attached graph of velocity vs time ...

shaded area above the x-axis = displacement from A to B

shaded area below the x-axis = displacement from B to A

the two shaded areas sum to zero, indicating a return to the starting point, A

View attachment 11188

- Mar 1, 2012

- 938

$\dfrac{T}{12}\left(\dfrac{3T^3}{4} - \dfrac{16T^2}{3} - 5T+ 60\right) = 0$

ignored the T/12 factor & multiplied the terms inside the parentheses by 12 to clear the fractions …

$9T^3 - 64T^2 -60T+720 = 0$

I graphed the cubic on my calculator and found T = 6 was a zero, then used synthetic division to find the quadratic factor …

$(T-6)(9T^2-10T-120) = 0$

$T = \dfrac{10 \pm \sqrt{4420}}{18}$

discarding the negative value for T …

$T = \dfrac{10+2\sqrt{1105}}{18} = \dfrac{5+\sqrt{1105}}{9}$

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- #6

Thank you so much!!

$\dfrac{T}{12}\left(\dfrac{3T^3}{4} - \dfrac{16T^2}{3} - 5T+ 60\right) = 0$

ignored the T/12 factor & multiplied the terms inside the parentheses by 12 to clear the fractions …

$9T^3 - 64T^2 -60T+720 = 0$

I graphed the cubic on my calculator and found T = 6 was a zero, then used synthetic division to find the quadratic factor …

$(T-6)(9T^2-10T-120) = 0$

$T = \dfrac{10 \pm \sqrt{4420}}{18}$

discarding the negative value for T …

$T = \dfrac{10+2\sqrt{1105}}{18} = \dfrac{5+\sqrt{1105}}{9}$