Mechanics- General motion in a straight line.

In summary, the conversation discusses a particle's motion along the X-axis with an acceleration of 6t-c. It is initially stationary and stops again at a point with x-coordinate of -4. The value of c is found by solving a quadratic equation using the equations for velocity and position when the particle is stationary.
  • #1
Shah 72
MHB
274
0
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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  • #2
What have you been able to do so far?

-Dan
 
  • #3
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
 
  • #4
topsquark said:
What have you been able to do so far?

-Dan
I got c=12 not -12
 
  • #5
Shah 72 said:
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
 
  • #6
topsquark said:
\(\displaystyle a = \dfrac{dv}{dt} = 6t - c\)

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
 
  • #7
topsquark said:
\(\displaystyle a = \dfrac{dv}{dt} = 6t - c\)

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
 
  • #8
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
\(\displaystyle 0 = 3t^2 - ct\)
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
 
  • #9
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
\(\displaystyle 0 = 3t^2 - ct\)
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!
 

Related to Mechanics- General motion in a straight line.

What is mechanics?

Mechanics is a branch of physics that deals with the study of motion and the forces that cause motion.

What is general motion in a straight line?

General motion in a straight line refers to the movement of an object along a straight path, where the object's speed and direction can change.

What is the difference between speed and velocity?

Speed is a measure of how fast an object is moving, while velocity is a measure of both speed and direction of motion.

What is acceleration?

Acceleration is the rate at which an object's velocity changes over time. It can be positive (speeding up), negative (slowing down), or zero (constant velocity).

What are the equations used to calculate motion in a straight line?

The equations used to calculate motion in a straight line are the equations of motion, which include the equations for displacement, velocity, acceleration, and time.

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