Question

A small particle starts from rest from the origin of an xy-coordinate system and travels in the xy-plane. Its acceleration in the x-direction is 2m/s^2, and its acceleration in the y-direction is 1m/s^2. What is the x-coordinate of the particle when the y-coordinate is 12m?

Answer 1

Answer:

When the y-coordinate is 12m, the x-coordinate of the particle is 24 m

Explanation:

Given;

y - component of acceleration = 1 m/s²

X - component of acceleration = 2 m/s²

distance traveled in y - direction, Dy = 12 m

To determine the distance traveled in X- direction, we obtain the duration of the displacement in y- direction.

Applying equation of motion;

Dy = ¹/₂ x at²

$t = \sqrt{\frac{2y}{a}} = \sqrt{\frac{2*12}{1}} = 4.9 s$

For distance traveled in x -direction;

Dx = ¹/₂ x at²

Dx = ¹/₂ x 2 x (4.9)² = 24.01 m ≅ 24m

Therefore, when the y-coordinate is 12m, the x-coordinate of the particle is 24 m

Answer 2

Answer:

24m

Explanation:

Consider one of the equations of motion as follows;

s = ut + $\frac{1}{2}$at²            ----------------------------(i)

Where;

s = vertical/horizontal displacement of the body in motion

u = initial vertical/horizontal displacement of the body

t = time taken for the displacement

a = vertical/horizontal acceleration of the body.

Now, since the particle being considered moves in an xy-coordinate system, then equation (i) above can be resolved into the x (horizontal) and y (vertical) components as follows;

Horizontal (x-coordinate) component

$s_{x}$ = $u_{x}$ t + $\frac{1}{2}$ $a_{x}$t²     ------------------(ii)

Where;

$s_{x}$ = horizontal displacement (x-coordinate) of the particle in motion

$u_{x}$ = initial horizontal displacement of the particle

t = time taken for the displacement

$a_{x}$ = horizontal (x-direction) acceleration of the body.

Vertical (y-coordinate) component

$s_{y}$ = $u_{y}$ t + $\frac{1}{2}$ $a_{y}$t²      -------------------(iii)

Where;

$s_{y}$ = vertical displacement (y-coordinate) of the particle in motion

$u_{y}$ = initial vertical displacement of the particle

t = time taken for the displacement

$a_{y}$ = vertical (y-direction) acceleration of the body.

(A) Now, using equation (iii), from the question;

$u_{y}$ = 0              [since the particle starts from rest, initial velocity is zero]

$a_{y}$ = 1m/s²        [acceleration in the y-direction]

$s_{y}$ = 12m           [y-coordinate value]

Substitute these values into equation (iii) as follows;

12 = 0 t +  $\frac{1}{2}$ (1) t²

12 =   $\frac{1}{2}$ t²          [Multiply through by 2]

24 = t²              [Solve for t]

t = $\sqrt{24}$ seconds

(B) Also, using equation (ii), from the question;

$u_{x}$ = 0              [since the particle starts from rest, initial velocity is zero]

$a_{x}$ = 2m/s²       [acceleration in the x-direction]

$s_{x}$ = ?               [x-coordinate value]

Substitute these values into equation (ii) as follows;

$s_{x}$ = 0 t +  $\frac{1}{2}$ (2) t²

$s_{x}$ =  t²             -------------------(iv)

But t = $\sqrt{24}$ seconds as calculated above, substitute this value into equation (iv)

$s_{x}$ = ($\sqrt{24}$)²              [Solve for $s_{x}$]

$s_{x}$ = 24

Therefore, the x-coordinate of the particle when the y-coordinate is 12m is 24m