# Use kinematic equations if acceleration is time dependent?

1. Feb 26, 2016

### fog37

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

2. Feb 26, 2016

### Samy_A

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

3. Feb 28, 2016

### fog37

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?

4. Feb 28, 2016

### ZapperZ

Staff Emeritus
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.

5. Feb 28, 2016

### fog37

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,