Physics, 27.06.2019 02:30 marianaescamilla404

Aparticle leaves the origin with an initial velocity of 4.90 m/s in the x direction, and moves with constant acceleration ax = -1.50 m/s^2 and ay = 3.00 m/s^2. how far does the particle move in the x direction before turning around?

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Answer from: mimi8734

it will move by d = 8.00 m in x direction before it will turn back

Explanation:

Here the initial velocity of particle is along +x direction

it is given as

v_i = 4.90 m/s

now its acceleration is given as

a_x = -1.50 m/s^2

a_y = 3.00 m/s^2

now when it turns back then the velocity in x direction will become zero

so we will say

v_f^2 - v_i^2 = 2 a d

0 - 4.90^2 = 2(-1.50) d

d = 8.00 m

so it will move by d = 8.00 m in x direction before it will turn back

Answer from: Quest

the independent variable, or the one that is intentionally manipulated is the speed.

Answer from: Quest

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Aparticle leaves the origin with an initial velocity of 4.90 m/s in the x direction, and moves with...

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