# Mechanics- General motion in a straight line.

• MHB
• Shah 72
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.
Shah 72
MHB
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.

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

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

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

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

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

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!

## 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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