Mechanics- General motion in a straight line.

[Shah 72]

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.

 

[topsquark]

What have you been able to do so far?

-Dan

 

[Shah 72]

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

 

[Shah 72]

What have you been able to do so far?

-Dan

I got c=12 not -12

 

[topsquark]

I got c=12 not -12

\(\displaystyle a = \dfrac{dv}{dt} = 6t - c\)

so \(\displaystyle v = \int_0^t (6t - c) ~ dt = 3t^2 - ct\)

\(\displaystyle v = \dfrac{dx}{dt} = 3t^2 - ct\)

Try to finish it.

-Dan

 

[Shah 72]

\(\displaystyle a = \dfrac{dv}{dt} = 6t - c\)

so \(\displaystyle v = \int_0^t (6t - c) ~ dt = 3t^2 - ct\)

\(\displaystyle v = \dfrac{dx}{dt} = 3t^2 - ct\)

Try to finish it.

-Dan

You mean I do s=t^3-ct^2/2

 

[Shah 72]

\(\displaystyle a = \dfrac{dv}{dt} = 6t - c\)

so \(\displaystyle v = \int_0^t (6t - c) ~ dt = 3t^2 - ct\)

\(\displaystyle v = \dfrac{dx}{dt} = 3t^2 - ct\)

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

 

[topsquark]

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
\(\displaystyle 0 = 3t^2 - ct\)
and
\(\displaystyle -4 = t^3 - \dfrac{1}{2} c t^2\)

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

-Dan

 

[Shah 72]

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
\(\displaystyle 0 = 3t^2 - ct\)
and
\(\displaystyle -4 = t^3 - \dfrac{1}{2} c t^2\)

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

-Dan

Thank you so much!