Question

A particle leaves the origin with an initial velocity and a constant acceleration . When the particle reaches its maximum x coordinate, what are

(a) its velocity,
(b) its position vector?

Answer 1

Answer:

(a)

$\displaystyle \vec v_f=<0,\frac{vy_o.ax-vx_o.ay}{ax}>$

(b)

$\displaystyle \vec r=<-\frac{vx_o^2}{2ax},\frac{vx_o}{2a^2x}\left(vx_o.ay-2ax.vy_o\right)>$

Explanation:

Constant Acceleration Motion

This describes a situation where an object is constantly changing its velocity over time, at a constant rate. If we treat the magnitudes as scalars (one fixed direction) then we have the following expression to find the final speed $v_f$ in term of the initial speed $v_o$, the constant acceleration a and the time t

$v_f=v_o+at$

The distance the object travels can also be found in terms of the same magnitudes

$\displaystyle x=v_ot+\frac{at^2}{2}$

If the object is moving in a two-dimension space, then all the magintudes are vectors, and the formulas change to

$\vec v_f=\vec v_o+\vec at$

$\displaystyle \vec r=\vec v_ot+\frac{\vec at^2}{2}$

The initial components of $\vec r$ are zero, as indicated in the question, so they are not included in the above formula.

The velocities and acceleration have two components in the x and y coordinates

The question does not clearly specify some conditions of the situation, so we'll assume some minimum facts to make this have sense

Assuming $vx_o$ is positive, if the object will eventually reach a maximum x-coordinate, then the x-component of the acceleration must be negative, so the x-component of the velocity decreases to zero and the object moves back in that direction. The acceleration will be written as

$\vec a=<ax,ay>$

Where ax is negative

The velocity can be written in vectorial notation as

$\vec v_f=<vx_o+ax.t,vy_o+ay.t>$

Where $vx_o$ and $vy_o$ are the initial velocity components

We can find the time $t_m$ where the object starts to change its direction in the x-axis by equating the $v_f$ component to zero

$vx_o+ax.t_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{vx_f-vx_o}{a}=\frac{0-vx_o}{ax}$

$\displaystyle t_m=-\frac{vx_o}{ax}$

Since ax is negative, $t_m$ is guaranteed to be positive.

(a)

The velocity can be computed at this specific time as

$\vec v_f=<vx_o+ax.t_m,vy_o+ay.t_m>$

Replacing $t_m$

$\vec v_f=<vx_o+ax.\left ( -\frac{vx_o}{ax} \right ),vy_o+ay.\left ( -\frac{vx_o}{ax} \right )>$

Simplifying

$\displaystyle \vec v_f=<0,\frac{vy_o.ax-vx_o.ay}{ax}>$

(b)

The postition vector at this specific time will be

$\displaystyle \vec r=\vec v_ot_m+\frac{\vec at_m^2}{2}$

$\displaystyle \vec r=<vx_o,vy_o>.\left ( -\frac{vx_o}{a} \right )+\frac{<ax,ay>.\left ( -\frac{vx_o}{a} \right )^2}{2}$

Operating

$\displaystyle \vec r=<-\frac{vx_o^2}{ax},-\frac{vx_o.vy_o}{ax}>+<\frac{vx_o^2}{2ax},\frac{vx_o^2ay}{2a^2x}>$

Simplifying

$\displaystyle \vec r=<-\frac{vx_o^2}{2ax},\frac{vx_o}{2a^2x}\left(vx_o.ay-2ax.vy_o\right)>$

Note: Assuming $vx_o$ was positive doesn't make our formulas less valid. If $vx_o$ was negative, then ax must be positive