- #1

Old_sm0key

- 17

- 0

- 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.

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.

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.

**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.