Use kinematic equations if acceleration is time dependent?

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Kinematic equations for constant acceleration cannot be directly applied when acceleration is time-dependent. Instead, one must integrate the time-dependent acceleration function to obtain velocity, and then integrate velocity to find position. The discussion also touches on the distinction between distance and displacement, clarifying that displacement is a vector quantity. Additionally, the concept of parameterization is introduced, highlighting that relationships can exist between different physical quantities. Understanding these principles is essential for accurately analyzing motion with variable acceleration.
fog37
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Hello forum,
The kinematic equations for motion with constant acceleration are:

v_f = v_0 + a*t
x_f = x_0 + v_0 * t +(0.5) a*t^2

The acceleration a is a constant.

Is it possible to use them if the acceleration is not constant but a function of time? For example, a(t)= 3t^2+2?
Can we simply replace a(t) in the equations above? I don't think so.

Do we need to solve dv/dt= a(t) for v(t) and the integrate v(t) to find the expression for x(t)?

I found a website that discusses time-dependent acceleration:
http://hyperphysics.phy-astr.gsu.edu/hbase/avari.html#c1

thanks
fog37
 
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fog37 said:
Hello forum,
The kinematic equations for motion with constant acceleration are:

v_f = v_0 + a*t
x_f = x_0 + v_0 * t +(0.5) a*t^2

The acceleration a is a constant.

Is it possible to use them if the acceleration is not constant but a function of time? For example, a(t)= 3t^2+2?
Can we simply replace a(t) in the equations above? I don't think so.

Do we need to solve dv/dt= a(t) for v(t) and the integrate v(t) to find the expression for x(t)?

I found a website that discusses time-dependent acceleration:
http://hyperphysics.phy-astr.gsu.edu/hbase/avari.html#c1

thanks
fog37
You can't just replace a by a(t) in the equations that are valid for constant acceleration.
Basically, you have to integrate a(t) to get v(t), and integrate v(t) to get x(t).

The website you mentioned shows it clearly: http://hyperphysics.phy-astr.gsu.edu/hbase/acons.html#c3
acc.jpg
 
Ok thanks! all clear.

We are usually working with position versus time, acceleration versus time, velocity versus time: x(t), a(t), v(t)

The distance traveled s is a scalar. Is it possible to express position x, acceleration a, and velocity v as a function of s? Is that called parametrization in term of arch length? How could I do that? Do you have a simple example?
 
fog37 said:
Ok thanks! all clear.

We are usually working with position versus time, acceleration versus time, velocity versus time: x(t), a(t), v(t)

The distance traveled s is a scalar. Is it possible to express position x, acceleration a, and velocity v as a function of s? Is that called parametrization in term of arch length? How could I do that? Do you have a simple example?

Wait.. back off a bit. "Distance traveled" is often designated as "displacement". This is NOT a scalar. It has distance AND direction!

Now, it is often treated as a "scalar" when one is only dealing with 1D problems. In that case, the only thing you care about if it is to the "left" or to the "right" of the origin. But do not confuse this as generalizing displacement as being a scalar.

Secondly, you can parameterize anything IF there is a relationship to connect them. In Hooke's law, the force can be expressed as a function of the displacement from equilibrium, i.e. a function of distance, not time.

Zz.
 
I did not mean displacement (which is a vector), either instantaneous or average, but the length of the trajectory at a specific instant of time as the independent variable for either velocity vector or position vector

Thanks,
 

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