(a)
![\displaystyle \vec v_f=](/tpl/images/0435/5631/77d9e.png)
(b)
![\displaystyle \vec r=](/tpl/images/0435/5631/ef15c.png)
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
in term of the initial speed
, the constant acceleration a and the time t
![v_f=v_o+at](/tpl/images/0435/5631/83fb4.png)
The distance the object travels can also be found in terms of the same magnitudes
![\displaystyle x=v_ot+\frac{at^2}{2}](/tpl/images/0435/5631/0a6c8.png)
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](/tpl/images/0435/5631/6fd37.png)
![\displaystyle \vec r=\vec v_ot+\frac{\vec at^2}{2}](/tpl/images/0435/5631/3d331.png)
The initial components of
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
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=](/tpl/images/0435/5631/58b2c.png)
Where ax is negative
The velocity can be written in vectorial notation as
![\vec v_f=](/tpl/images/0435/5631/2c05b.png)
Where
and
are the initial velocity components
We can find the time
where the object starts to change its direction in the x-axis by equating the
component to zero
![vx_o+ax.t_m=0](/tpl/images/0435/5631/b41a2.png)
Solving for ![t_m](/tpl/images/0435/5631/2b572.png)
![\displaystyle t_m=\frac{vx_f-vx_o}{a}=\frac{0-vx_o}{ax}](/tpl/images/0435/5631/4bae4.png)
![\displaystyle t_m=-\frac{vx_o}{ax}](/tpl/images/0435/5631/c1c80.png)
Since ax is negative,
is guaranteed to be positive.
(a)
The velocity can be computed at this specific time as
![\vec v_f=](/tpl/images/0435/5631/d8c9f.png)
Replacing ![t_m](/tpl/images/0435/5631/2b572.png)
![\vec v_f=](/tpl/images/0435/5631/5c340.png)
Simplifying
![\displaystyle \vec v_f=](/tpl/images/0435/5631/77d9e.png)
(b)
The postition vector at this specific time will be
![\displaystyle \vec r=\vec v_ot_m+\frac{\vec at_m^2}{2}](/tpl/images/0435/5631/7375c.png)
![\displaystyle \vec r=.\left ( -\frac{vx_o}{a} \right )+\frac{.\left ( -\frac{vx_o}{a} \right )^2}{2}](/tpl/images/0435/5631/a7884.png)
Operating
![\displaystyle \vec r=+](/tpl/images/0435/5631/331c7.png)
Simplifying
![\displaystyle \vec r=](/tpl/images/0435/5631/ef15c.png)
Note: Assuming
was positive doesn't make our formulas less valid. If
was negative, then ax must be positive