Irodov 1.24: One Parabola, Constant Acceleration Underneath

Problem Irodov 1.24

A point $A$ has position vector $\vect r(t) = at\,\vect i - bt^2\,\vect j$ relative to a fixed origin, where $a$ and $b$ are positive constants and $\vect i$ , $\vect j$ are the unit vectors of the $x$ and $y$ axes. Find:

the trajectory’s equation $y(x)$ , and sketch it
the velocity $\vect v(t)$ and acceleration $\vect w(t)$ , and the time dependence of their moduli $v(t)$ and $w(t)$
the time dependence of the angle $\alpha$ between $\vect w$ and $\vect v$
the mean velocity vector $\langle\vect v\rangle$ averaged over the first $t$ seconds, and its modulus

Given

$\vect r(t)=at\vect i-bt^2\vect j$

Find

(a)

y(x)

, (b)

\vect v(t)

\vect w(t)

v(t)

w(t)

, (c)

\alpha(t)

, (d)

\langle\vect v\rangle

and its modulus

Solution

Step 1Eliminate

t

for the trajectory

The two scalar components of $\vect r(t)$ are $x=at$ and $y=-bt^2$ . Solving the first for $t$ and substituting into the second removes the parameter.

\begin{align*}t &= \frac{x}{a} \\ y &= -b\left(\frac{x}{a}\right)^2\end{align*}

y = -\dfrac{b}{a^2}x^2

(1)

a parabola through the origin, opening in the direction of $-\vect j$ .

Step 2Velocity and acceleration by differentiation

Differentiating $\vect r(t)$ once gives $\vect v(t)$ and a second time gives $\vect w(t)$ ; each modulus is the root sum of squares of its own components.

\begin{align*}\vect v(t) &= \frac{d\vect r}{dt} = \frac{d}{dt}\bigl(at\vect i - bt^2\vect j\bigr) \\ \vect w(t) &= \frac{d\vect v}{dt} = \frac{d}{dt}\bigl(a\vect i - 2bt\vect j\bigr)\end{align*}

\vect v(t)=a\vect i-2bt\vect j,\qquad \vect w=-2b\vect j

(2)

\begin{align*}v(t) &= \sqrt{v_x^2+v_y^2} = \sqrt{a^2+(2bt)^2} \\ w(t) &= \sqrt{w_x^2+w_y^2} = \sqrt{0^2+(-2b)^2}\end{align*}

v(t)=\sqrt{a^2+4b^2t^2},\qquad w=2b

(3)

$\vect w$ is constant, always along $-\vect j$ .

Step 3The angle between

\vect w

and

\vect v

from their cross and dot products

With both vectors written in components, the sine and cosine of the angle between them come from the cross product’s $z$ component and the dot product, and their ratio isolates $\tan\alpha$ without needing either magnitude separately.

\begin{align*}\vect v\times\vect w &= (a\vect i-2bt\vect j)\times(-2b\vect j) = -2ab\,(\vect i\times\vect j) = -2ab\,\vect k \\ \vect v\cdot\vect w &= (a\vect i-2bt\vect j)\cdot(-2b\vect j) = 4b^2t \\ \tan\alpha &= \frac{|\vect v\times\vect w|}{\vect v\cdot\vect w} = \frac{2ab}{4b^2t}\end{align*}

\tan\alpha = \dfrac{a}{2bt}

(4)

The trajectory y=-bx^2/a^2, a parabola through the origin opening along - j, the part (a) plot. At the point A the velocity v runs along the tangent and the acceleration w along - j; the wedge between them is α. — The trajectory $y=-bx^2/a^2$ , a parabola through the origin opening along $-\vect j$ , the part (a) plot. At the point $A$ the velocity $\vect v$ runs along the tangent and the acceleration $\vect w$ along $-\vect j$ ; the wedge between them is $\alpha$ .

Step 4Mean velocity over the first

t

seconds

Dividing the position vector by the elapsed time gives the average, and its modulus follows the same way as in Step 2.

\begin{align*}\langle\vect v\rangle &= \frac{\vect r(t)}{t} = \frac{at\vect i-bt^2\vect j}{t} \\ &= a\vect i-bt\vect j\end{align*}

\langle\vect v\rangle = a\vect i-bt\vect j,\qquad |\langle\vect v\rangle| = \sqrt{a^2+b^2t^2}

(5)

Check

At $t=0$ , Step 2 gives $\vect v(0)=a\vect i$ , purely horizontal, and Step 4 gives $\langle\vect v\rangle\to a\vect i$ as $t\to0$ : the two must agree at the start, and they do. Also $\tan\alpha\to\infty$ as $t\to0$ , i.e. $\alpha\to90^\circ$ , consistent with $\vect v(0)\perp\vect w$ since $\vect v(0)$ is purely along $\vect i$ and $\vect w$ is purely along $\vect j$ .

Remark

Because $\vect w$ never changes, this is a constant-acceleration motion exactly like projectile motion, with $-\vect j$ playing the role of the downward vertical and $a\vect i$ the role of a constant horizontal launch speed; the parabola $y=-bx^2/a^2$ is the same shape gravity draws, only here nothing physical is falling; it is simply how a linear-in- $t$ and a quadratic-in- $t$ component combine.