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dmitriy555 [2]
3 years ago
12

A small particle starts from rest from the origin of an xy-coordinate system and travels in the xy-plane. Its acceleration in th

e 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?
Physics
2 answers:
Oduvanchick [21]3 years ago
8 0
<h2>Answer:</h2>

24m

<h2>Explanation:</h2>

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;

<u>Horizontal (x-coordinate) component</u>

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.

<u>Vertical (y-coordinate) component</u>

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]

<em>Substitute these values into equation (iii) as follows;</em>

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]

<em>Substitute these values into equation (ii) as follows;</em>

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

Gennadij [26K]3 years ago
4 0

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

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