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Physics Mechanics- General motion in a straight line.

Shah 72

Member
Apr 14, 2021
193
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

Well-known member
MHB Math Helper
Aug 30, 2012
1,207
What have you been able to do so far?

-Dan
 

Shah 72

Member
Apr 14, 2021
193
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

Member
Apr 14, 2021
193

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,207
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

Member
Apr 14, 2021
193
\(\displaystyle a = \dfrac{dv}{dt} = 6t - c\)

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

Member
Apr 14, 2021
193
\(\displaystyle a = \dfrac{dv}{dt} = 6t - c\)

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

Well-known member
MHB Math Helper
Aug 30, 2012
1,207
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

Member
Apr 14, 2021
193
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!!