# Plotting displacement v time and velocity v time for a bouncing particle

Old_sm0key
Homework Statement:
Not a homework question per se. I revisited introductory kinematics, as taught at British A Level, and am struggling to reproduce the stadnard plots for displacement and velocity versus time.
Relevant Equations:
The standard 1D kinematics equations plus introductory differentiation
Initial displacement is h above the ground ie ##s\left ( t =0\right )=h##. I've chosen the ground as the vertical origin with upwards as the positive direction. Gravity will therefore always act in negative direction throughout. Here are the graphs I which to reproduce from first principles, where the motion starts at the red vertical line owing to my initial conditions: ##0\leqslant t< A##
Starting from ##v=u+at## we derive displacement thus:
$$\int_{0}^{t} vdt^\prime=\int_{0}^{t}\left ( u+at \right )dt^\prime$$
$$s\left ( t \right )-h=ut+\frac{1}{2}at^2$$
Using conditions ##t=0,u=0, a=-g\, \forall t \to s\left ( t \right )=h-\frac{1}{2}gt^2## which would give a parabola for first portion of displacement graph. So happy.
The corresponding velocity graph is ##\frac{\mathrm{d} s}{\mathrm{d} t}=-gt##, so giving a negative straight line. Happy.

##A\leqslant t< B##
Now I'm stuck with the displacement curve because I cannot exploit ##t=0## on the integral limits to simplify the maths...?
Plus it seems a fudge to suddenly invoke ##s\left ( t \right )=vt-\frac{1}{2}at^2## that I see in some textbooks.

Homework Helper
Gold Member
If you want to avoid the fudge of the discontinuity of the velocity at the bounce, you can model the ball as having a spring attached to its bottom. This will round off the curve with a sinusoidal and give it a collision duration of half a period. You can sharpen it by increasing the spring constant ##k.## so that the period of oscillations is much much smaller than the time between bounces.

Homework Helper
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With what are you trying to reproduce them? If in a spreadsheet, it's just a matter of first converting t to be in the range 0 to A using MOD.

Old_sm0key
Thanks both, however I probs didn't make myself too clear.

(I accept the velocity discontinuity @ bounce ie precluding differentiation)

I want to derive the s vs t equation, just like I did above, now for ##A\leqslant t< B## to produce the shape shown in plot. Just can't get the right s(t) function..!

Homework Helper
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2022 Award
$$\int_{0}^{t} vdt^\prime=\int_{0}^{t}\left ( u+at \right )dt^\prime$$
In order to correctly distinguish between your integration variable and the limits, this should be$$\int_{0}^{t} vdt^\prime=\int_{0}^{t}\left ( u+at^\prime \right )dt^\prime$$If I’ve understood what you are asking…

Let T be the duration of each ‘cycle’ (from t=0 to A). This is a constant you can express in terms of h and g.

You have chosen to make B halfway through the 2nd cycle. Not sure if that’s deliberate, but no matter.

You can use the method you used for the 1st cycle - but change the integration limits: from T to t+T/2. Because the process is cyclic, your initial conditions at the start of each cycle are identical to those at t=0.

Staff Emeritus
Homework Helper
Gold Member
Thanks both, however I probs didn't make myself too clear.

(I accept the velocity discontinuity @ bounce ie precluding differentiation)

I want to derive the s vs t equation, just like I did above, now for ##A\leqslant t< B## to produce the shape shown in plot. Just can't get the right s(t) function..!
Why not include B in the interval? I.e. ##\ A\le t \le B \ ##
You can actually go father, to the next zero of ##s(t)##, maybe call it ##C##.

At ##t=B## you have ##s(t) = h## and ##v(t) = 0##.

Either use indefinite integrals and evaluate the constants of integration according to the above values (at ##t=B## ) ,

or use definite integrals and integrate from ##t## to ##B##, or from ##B## to ##t##. Just be consistent with the direction of the integration.