MHB General motion in a straight line.

[Shah 72]

In q(c) I calculated the distance using t= 4.25 S . I get S=-3.23×10^-3
I added 5.78m to this which is the distance between A and B, I get 5.78m. Is this the correct method to prove that the bird returns to A. Pls help

 

[skeeter]

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

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

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

 

[Shah 72]

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

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

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

https://www.physicsforums.com/attachments/11188

Thanks so much!

 

[Shah 72]

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

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

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

https://www.physicsforums.com/attachments/11188

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

https://www.physicsforums.com/attachments/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).

 

[skeeter]

I integrated the velocity function and evaluated it from 0 to T and set the result equal to zero.

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

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 …

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

 

[Shah 72]

I integrated the velocity function and evaluated it from 0 to T and set the result equal to zero.

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

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 …

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

Thank you so much!