Irodov 1.28: Chords Are Mean Velocities in Disguise – Amit & Deepali Sinha Foundation

Problem Irodov 1.28

A small body is launched with an initial velocity vector $\vect v_0$ at some angle to the horizontal and moves freely under a constant gravitational acceleration $\vect g$ , air resistance neglected. Find:

the body’s displacement $\vect r(t)$ as a function of the elapsed time $t$
the mean velocity vector $\langle\vect v\rangle$ , averaged over the first $t$ seconds of the flight, and also averaged over the whole flight, from launch until the body returns to its launch height

Given

$\vect v_0$
$\vect g$

Find

(a)

\vect r(t)

, (b)

\langle\vect v\rangle_t

over the first

t

seconds and

\langle\vect v\rangle

over the whole flight

Solution

Step 1Velocity and displacement under constant

\vect g

With acceleration $\vect w=\vect g$ constant and the launch conditions $\vect v(0)=\vect v_0$ , $\vect r(0)=\vect 0$ , one integration gives the velocity at any time and a second gives the position.

\vect v(t) = \vect v_0+\int_0^t \vect g\,d\tau = \vect v_0+\vect g t \tag{1}

\begin{align*}\vect r(t) &= \int_0^t \vect v(\tau)\,d\tau \\ &= \int_0^t\bigl(\vect v_0+\vect g\tau\bigr)\,d\tau \\ &= \left.\left(\vect v_0\tau+\dfrac{\vect g\tau^2}{2}\right)\right|_0^t\end{align*}

\vect r(t) = \vect v_0 t+\dfrac{\vect g t^2}{2}

(2)

Step 2Mean velocity over the first

t

seconds

Dividing the displacement (2) by the elapsed time gives the mean velocity over $[0,t]$ .

\begin{align*}\langle\vect v\rangle_t &= \dfrac{\vect r(t)}{t} \\ &= \dfrac{\vect v_0 t+\dfrac{\vect g t^2}{2}}{t}\end{align*}

\langle\vect v\rangle_t = \vect v_0+\dfrac{\vect g t}{2}

(3)

exactly the average of the initial and the instantaneous velocity at $t$ , since both $\vect v(t)$ and $\vect r(t)/t$ are linear in $t$ under constant acceleration.

The same flight with the chord from O to P drawn on it. That chord is the displacement r(t), and the mean velocity over the first t seconds points along it rather than along the tangent at P. — The same flight with the chord from $O$ to $P$ drawn on it. That chord is the displacement $\vect r(t)$ , and the mean velocity over the first $t$ seconds points along it rather than along the tangent at $P$ .

Step 3The flight time from the landing condition

The flight time $T$ is the instant the body returns to launch height, so the displacement then has no component along $\vect g$ . Dotting the displacement (2) with $\vect g$ and keeping the nonzero root isolates $T$ .

\begin{align*}0 &= \vect r(T)\cdot\vect g \\ &= \left(\vect v_0 T+\dfrac{\vect g T^2}{2}\right)\cdot\vect g \\ &= T(\vect v_0\cdot\vect g)+\dfrac{T^2 g^2}{2} \\ &= T\left[(\vect v_0\cdot\vect g)+\dfrac{T g^2}{2}\right]\end{align*}

$T=0$ is the launch instant; the landing is the other root, where the bracket vanishes.

(\vect v_0\cdot\vect g)+\dfrac{T g^2}{2} = 0

T = -\dfrac{2(\vect v_0\cdot\vect g)}{g^2}

(4)

Step 4Mean velocity over the whole flight

The whole-flight average is (3) evaluated at $t=T$ ; substituting the flight time (4) and simplifying gives the result.

\begin{align*}\langle\vect v\rangle &= \vect v_0+\dfrac{\vect g T}{2} \\ &= \vect v_0+\dfrac{\vect g}{2}\left(-\dfrac{2(\vect v_0\cdot\vect g)}{g^2}\right)\end{align*}

\langle\vect v\rangle = \vect v_0-\dfrac{\vect g(\vect v_0\cdot\vect g)}{g^2}

(5)

which is $\vect v_0$ with its component along $\vect g$ removed, i.e. exactly the horizontal part of $\vect v_0$ .

Check

$\langle\vect v\rangle$ has zero component along $\vect g$ by construction: dotting the boxed result with $\vect g$ gives $\vect v_0\cdot\vect g-g^2(\vect v_0\cdot\vect g)/g^2=0$ , consistent with the whole-flight chord, which runs from $O$ to a landing point at the same height and is therefore purely horizontal. And as $t\to0$ , Step 2’s $\langle\vect v\rangle_t\to\vect v_0$ , the instantaneous launch velocity, as any short-time average must.

Remark

Because $\vect v(t)$ is linear in $t$ , the mean of $\vect v$ over any interval equals its value at the interval’s midpoint. Here that midpoint is $t=T/2=-(\vect v_0\cdot\vect g)/g^2$ , which (1) gives as $\vect v_0-\vect g(\vect v_0\cdot\vect g)/g^2$ , exactly the whole-flight average above: the mean velocity of a parabolic flight equals the velocity at the apex, where the component of $\vect v$ along $\vect g$ momentarily vanishes.