MHB Mechanics- General motion in a straight line.

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A particle moves along the X-axis with acceleration defined as a = 6t - c, starting from the origin and initially at rest. The particle returns to a stationary state at x = -4, prompting a discussion on finding the constant c. Participants explore integrating the acceleration to find velocity and position equations, leading to the conclusion that both velocity and position must equal zero at the stationary points. The correct approach involves solving the equations derived from setting velocity to zero and substituting into the position equation. Ultimately, the value of c is determined to be 6.
Shah 72
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A particle starts at the origin and moves along the X- axis. The acceleration of the particle in the direction of the positive x-axis is a= 6t-c for some constant c. The particle is initially stationary and it is stationary again when it is at the point with x coordinate = -4. Find the value of c.
I don't understand how to calculate.
 
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What have you been able to do so far?

-Dan
 
topsquark said:
What have you been able to do so far?

-Dan
V= integration (6t-c) with limits 0 and -4. I got the ans c= -12. But it's wrong and the textbook ans is 6
 
topsquark said:
What have you been able to do so far?

-Dan
I got c=12 not -12
 
Shah 72 said:
I got c=12 not -12
[math]a = \dfrac{dv}{dt} = 6t - c[/math]

so [math]v = \int_0^t (6t - c) ~ dt = 3t^2 - ct[/math]

[math]v = \dfrac{dx}{dt} = 3t^2 - ct[/math]

Try to finish it.

-Dan
 
topsquark said:
[math]a = \dfrac{dv}{dt} = 6t - c[/math]

so [math]v = \int_0^t (6t - c) ~ dt = 3t^2 - ct[/math]

[math]v = \dfrac{dx}{dt} = 3t^2 - ct[/math]

Try to finish it.

-Dan
You mean I do s=t^3-ct^2/2
 
topsquark said:
[math]a = \dfrac{dv}{dt} = 6t - c[/math]

so [math]v = \int_0^t (6t - c) ~ dt = 3t^2 - ct[/math]

[math]v = \dfrac{dx}{dt} = 3t^2 - ct[/math]

Try to finish it.

-Dan
You take both the equations = 0 as stationary .
Then solve the quadratic equation by using b^2-4ac=0. I still don't get the ans
 
Shah 72 said:
You take both the equations = 0 as stationary .
Then solve the quadratic equation by using b^2-4ac=0. I still don't get the ans
Your formula for s (I've been calling it x) is correct. Now you know when the motion of the particle stops, v(t) = 0 and x(t) = -4. So
[math]0 = 3t^2 - ct[/math]
and
[math]-4 = t^3 - \dfrac{1}{2} c t^2[/math]

My advice is to solve the top equation for ct, multiply it by t (to get [math]ct^2[/math]) and then plug that into the bottom equation.

-Dan
 
topsquark said:
Your formula for s (I've been calling it x) is correct. Now you know when the motion of the particle stops, v(t) = 0 and x(t) = -4. So
[math]0 = 3t^2 - ct[/math]
and
[math]-4 = t^3 - \dfrac{1}{2} c t^2[/math]

My advice is to solve the top equation for ct, multiply it by t (to get [math]ct^2[/math]) and then plug that into the bottom equation.

-Dan
Thank you so much!
 

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