Calculate distance travelled by particle with positive velocity and negative acceleration in fixed time

Question

Let's assume a ball is thrown upwards with an initial velocity V and gravitational acceleration -A is acting on it in downward direction At some height H the particle with reach its max height and start falling down in negative direction I want to calculate the distance travelled by the ball and not the displacement

I can use s = ut + ½at² to calculate displacement but what i want is the distance travelled

For example, when the ball reaches the ground the displacement will be 0 but distance travelled will be twice the max height reached

How can I calculate distance travelled at any time t?

Answer 1

Consider a short time $\delta t$. The ball's velocity at this point is $v_0 - a t$, where $v_0 = v (t = 0)$, can be determined from simple kinematics. The distance travelled in this time is $\left| v_0 - a t \right| \delta t$. To find the distance traveled from $t = 0$ to some arbitrary $t$, simply integrate

$$d = \int_0^t \left| v_0 - a t' \right| d t'$$

There is no clean closed form for this. If we take $a$ and $v_0$ to be positive (downward acceleration, upwards initial velocity), then for $t \le \frac{v_0}{a}$ which corresponds to the upwards part of the motion, the distance will be identical to the displacement: $d (t) = v_0 t - \frac{1}{2} a t^2$. At $t_c = \frac{v_0}{a}$, the distance traveled is $d_c = \frac{v_0^2}{2 a}$. For $t > \frac{v_0}{a}$, the distance is $d_c$ plus the magnitude of the downward displacement. In particular, $d (t) = d_c + \frac{1}{2} a (t - t_c)^2$.

Answer 2

Let's make this a bit more general. Consider $s:[0,T]\rightarrow \mathbb{R}^n$, a parameterization of the trajectory the ball follows ($s(t)=$ position of the ball at time $t$, and $n=1$ in your case).

In a small fraction of time $dt$, your ball moves along the trajectory, approximately in a straight line tangent to the trajectory, by an amount $|g'(t)|dt$. Sum these amounts for all times to get your desired length: $$\int_0^T |s'(t)|dt$$

Answer 3

If you want a simple formula not involving integration, then try the following: First note that the particle reaches its maximum height $H$ where $$H=\frac{V^2}{2A}$$ at time $$T=\frac{V}{A}$$

(Taking the positive value of $A$, e.g. $9.8 ms^{-2}$)

So, for $t\leq T$, the distance travelled is $$s=Vt-\frac12At^2$$

For $t>T$, the total distance is $$2H-s$$ $$=\frac{V^2}{A}-Vt+\frac12At^2$$

Note that the second formula will still be valid when the particle descends to a level below its initial position.

I hope this helps.