accelerationhomework-and-exerciseskinematicsnewtonian-mechanics
I understand that when drawing an acceleration-time graph for a falling object (with initial velocity = 0), the y-intercept should be $9.81ms^{-2}$ and acceleration should end up being $0ms^{-2}$ as it reaches terminal velocity.
However, I do not understand why the rate of change of acceleration should be decreasing (as shown in the graph below).
Can someone explain to me why the rate of change of acceleration changes rather than the graph being a straight line?
Thanks in advance.
A simple example, particle moving in one dimension, say $x$:
A particle has a (negative) acceleration equal to $-2m/s^2$ (note it is constant acceleration, yet negative, i.e opposite to direction of motion or to direction of cummulant force applied to particle)
Another particle has a (decreasing) acceleration equal to $1/(kt)^2 m/s^2$ (note it is positive acceleration i.e in the same direction as direction of motion, yet decreasing in magnitude with time)
To sum up, negative or opposite acceleration refers to direction, while decreasing acceleration refers to magnitude.
The above are only illustrative examples, many more examples of negative and decreasing accelerations (or both).
Some mock v-t graphs for various types of acceleration:
The phrase freely falling really means freely moving, that is nothing is pushing or pulling the object while it's moving. So the object is freely falling even when it's rising! Like all things Physics has its own special terminology and I'm afraid you just have to get used to it.
Re the acceleration: the acceleration is the change of velocity with time. If I throw a stone upwards at 98.1 m/s then (ignoring air resistance) in the first second it slows by 9.81 m/s to 88.29 m/s, in the second second it slows by 9.81 m/s to 78.48 m/s and so on until after 10 seconds its velocity has slowed to zero. Each second it slows by the same 9.81 m/s, so the acceleration has the constant value of -9.81 m/sec$^2$ (the - sign means the acceleration is downwards).
Best Answer
Suppose that the drag force is proportional to the speed of the falling object then the equation of motion of the falling object is $ma = mg -kv \Rightarrow \dfrac {da}{dt} = - \dfrac k m a$.
So the slope of an acceleration against time graph is negative and gets less negative as the acceleration decreases.
A similar result can be obtained if the frictional force is proportional to the speed squared etc.