Irodov 1.25: Velocity Meets Acceleration at Forty-Five

Problem Irodov 1.25

A point moves in the plane $xy$ so that its coordinates vary with time $t$ as $x = at$ and $y = at(1-\alpha t)$ , where $a$ and $\alpha$ are positive constants. Find:

the equation of the point’s trajectory, giving $y$ as a function of $x$ ; plot this function
the point’s velocity $v$ and acceleration $w$ , each as a function of time
the instant $t_0$ at which the velocity vector and the acceleration vector make an angle of $\pi/4$ with each other

Given

$x=at$
$y=at(1-\alpha t)$

Find

(a)

y(x)

, (b)

v(t)

and

w(t)

, (c)

t_0

where the angle between

\vect v

and

\vect w

\pi/4

Solution

Step 1Eliminate

t

for the trajectory

The first coordinate gives the time in terms of the abscissa; substituting that into the second removes the parameter and leaves the path itself.

\begin{align*}t &= \frac{x}{a} \\ y &= a\cdot\frac{x}{a}\left(1-\alpha\frac{x}{a}\right) \\ &= x\left(1-\frac{\alpha}{a}x\right)\end{align*}

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

(1)

a downward parabola through the origin, coming back through $y=0$ at $x=a/\alpha$ .

Step 2Differentiate for velocity and acceleration

Componentwise,

v_x = \dot x = a,\qquad v_y = \dot y = a(1-2\alpha t), \tag{2}

w_x = \dot v_x = 0,\qquad w_y = \dot v_y = -2a\alpha. \tag{3}

So the acceleration is constant, pointing straight in the $-y$ direction, while the velocity’s $x$ part never changes and its $y$ part decays and reverses:

v = a\sqrt{1+(1-2\alpha t)^2},\qquad w = 2a\alpha = \text{const}

(4)

Step 3Find where

\vect v

and

\vect w

are at

\pi/4

Substituting the components (2) into that condition and working it down one operation at a time isolates $t_0$ .

\begin{align*}v_x &= -v_y \\ a &= -a(1-2\alpha t_0) \\ 1 &= -(1-2\alpha t_0) \\ 1 &= 2\alpha t_0-1 \\ 2\alpha t_0 &= 2\end{align*}

t_0 = \dfrac{1}{\alpha}

(5)

The trajectory y(x), drawn for a=α=1: a downward parabola through the origin, crossing back to y=0 at x=a/α. At that crossing, reached at t_0, the velocity v and the acceleration w are drawn with the angle π/4 between them. — The trajectory $y(x)$ , drawn for $a=\alpha=1$ : a downward parabola through the origin, crossing back to $y=0$ at $x=a/\alpha$ . At that crossing, reached at $t_0$ , the velocity $\vect v$ and the acceleration $\vect w$ are drawn with the angle $\pi/4$ between them.

Check

At $t_0=1/\alpha$ : $x_0=a/\alpha$ , and substituting into the trajectory (1) gives $y_0 = a/\alpha - (\alpha/a)(a/\alpha)^2 = 0$ , exactly the second crossing point marked in the figure, an independent confirmation that $t_0$ was placed correctly. There $v_y=a(1-2)=-a=-v_x$ , so $\vect v$ sits at $45^\circ$ below the $x$ axis while $\vect w$ points straight down, an angle of exactly $\pi/4$ between them.

Remark

The trajectory is exactly ordinary projectile motion in disguise: with constant $v_x=a$ and constant acceleration $w=2a\alpha$ along $-y$ , $x=at$ and $y=at(1-\alpha t)$ take the same shape $x=v_xt$ , $y=v_xt-wt^2/2$ as a projectile launched with both velocity components equal to $a$ at $t=0$ , i.e. at $45^\circ$ above the horizontal with speed $v_0=a\sqrt2$ , under a downward acceleration $2a\alpha$ , so the parabola, the decaying $v_y$ , and the constant $w$ are the familiar projectile story wearing different symbols.