# Acceleration-time graph for a falling object

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?

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.

• Sorry, can you correct my differentiation please? I keep getting a positive rate of change... $a=g-\frac{kv}{m}$ so, $a=g-\frac{ks}{m}\times t^{-1}$so, $\frac{da}{dt}=-1\times -\frac{ks}{m}\times t^{-2}$ so,$\frac{da}{dt}=\frac{k}{m}\times a$ Commented Nov 12, 2019 at 8:01
• @kimchiboy03 Don't write $v$ as $st^{-1}$. Commented Nov 12, 2019 at 8:26
• $v \ne s \times t^{-1}$ rather $v = \frac{ds}{dt}$ Commented Nov 12, 2019 at 8:28
• Oh, so I should do instead, $a=g-\frac{kv}{m}$ so, $\frac{da}{dt}=-\frac{k}{m}\times \frac{d^{2}s}{dt^{2}}$ so, $\frac{da}{dt}=-\frac{k}{m}\times \frac{dv}{dt}$ so, $\frac{da}{dt}=-\frac{k}{m}\times a$ Commented Nov 12, 2019 at 8:33

In an ideal scenario, yes it would be straight, but in this case, you have to take into account drag forces present. Drag force $$F_D=C_DA\frac{\rho v^2}{2}=kv^2$$, where $$k$$ is some constant. As such, for an object vertical falling motion, velocity increases, and that means drag force increases at an increasing rate. (Consider a $$y=x^2$$ graph).

Since acceleration of an object is given by $$\frac{F_W-F_D}{m}$$, with $$F_D$$ increasing at an increasing rate, acceleration would de decreasing at a decreasing rate rate.

Since you don't mention the medium that the object is falling in, one could assume the medium is air, but it also could be falling in a liquid. In either case, since air can be (is) treated as a fluid, the object is affected by drag, which is a type of friction or force acting opposite to the relative motion of the object and is proportional to the square of the velocity of the object.

The drag equation is $$F_{D}={\tfrac {1}{2}}\rho v^{2}C_{D}A$$,   where
$$F_{D}$$ is the drag force,
$$\rho$$ is the density of the fluid,
$$v$$ is the speed of the object relative to the fluid,
$$A$$ is the cross sectional area, and
$$C_{D}$$ is the drag coefficient (a dimensionless number).

The drag coefficient depends on other factors (the shape of the object and on the Reynolds number).

What this all means is that, as the object accelerates (i.e. speed increases), the force (drag) opposing the acceleration increases with the square of the speed, resulting in a non-linear nett acceleration over time.

Let's assume that the drag force is given as $$F_{drag}=-bv$$ then the net acceleration is given as $$a=g-\frac{bv}{m}$$ now differentiate this equation with respect to time then,$$\frac{da}{dt}=-\frac{ba}{m}$$ Now it is clear from this equation that the slope is negative and the curve falls according to the given equation:- $$a=exp(\frac{-bt}{m})+constant$$.