Question

What is the nature of the trajectory of a particle having a uniformly accelerated motion in a plane?

Answer 1

it travels in straight path it's velocity increases equal amounts in equal interval of time

Answer 2

Answer:

The path traced out is parabolic

According to the definition of the problem,

if the initial velocity is v0, then the distance that would be traveled in the direction, y, which is parallel to v0 is

y=(v0*t)

The distance, xp, traveled were there no initial velocity under constant acceleration is

xp=ap*t^2/2

However, ap has two components, one that is parallel to y and one that is perpendicular to y, which will be referred to as x. If xp makes an angle theta which y, then the two components are ay=ap*cos(theta) and ax=sin(theta)*ap

So,

x=ap*sin(theta)*t^2/2 (where the coordinate system is chosen so that of the initial value of of x is 0),

where as, in the y direction the distance traveled is

y=v0*t+ap*cos(theta)*t^2/2 (where the coordinate system is chosen so that of the initial value of of y is 0).

One can already see that both the motion in both the x and y direction are parabolic in the parameter t (the time).

However, continuing to show that the above two equations define a parabola in terms of x and y,

the above equation for

x=ap*sin(theta)*t^2/2 implies that

(1) t^2/2 = x*ap*sin(theta)

and

(2) t=sqrt(2*x*ap*sin(theta))

Substituting for t and t^2/2,

one gets,

y=v0*sqrt(2*x/(ap*sin(theta)))+ap*cos(theta)*x/(ap*sin(theta))

=v0*sqrt(2*x/(ap*sin(theta)))+cot(theta)*x

moving the last term to the right side of the equation, one gets,

y -cot(theta)*x=v0*sqrt(2*x/(ap*sin(theta))).

squaring both sizes, one gets,

y^2–2y*x*cot(theta)+(x^2)*(cot(theta))^2=(v0^2)*2*x/(ap*sin(theta)

This equation as the form of

y^2 +c1*x^2-c2*x*y-c3*x=0 (where c1, c2, and c3 are constants that depend on the acceleration, ap, initial velocity, v0, and the angle between the two, , theta), which is a parabola.

The parabola is simplified if one chooses the one of the x and y coordinates to be parallel to the direction of acceleration (as done by Paul Grimshaw), instead of along the direction of the initial velocity (as I did).

Explanation: