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23 October, 17:33

A particle leaves the origin with an initial velocity v → = (3.00iˆ) m/s and a constant acceleration a → = (-1.00iˆ - 0.500jˆ) m/s2. When it reaches its maximum x coordinate, what are its (a) velocity and (b) position vector?

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  1. 23 October, 17:53
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    the position vector (x, y) will be (1.5 m,-2.25 m) and the velocity vector (vx, vy) will be (0 m/s, - 1.5 m/s) when x reaches its maximum x coordinate

    Explanation:

    Since the velocity is related with the acceleration and coordinates through

    vx²=v₀x²+2*ax*x

    where

    vx = velocity in the x direction

    v₀x = initial velocity in the x direction = 3 m/s

    ax = acceleration in the x direction = - 1.00 m/s²

    x = coordinates in the x-axis

    when x reaches its maximum coordinate, then vx=0

    thus

    vx²=v₀x²+2*ax*x

    0 = (3 m/s) ² + 2 * (-1.00 m/s²) * x

    x = 1.5 m

    also for the time t

    vx = v₀x + ax*t → t = (vx-v₀x) / ax = (0 - 3 m/s) / (-1.00 m/s²) = 3 seconds

    for the y coordinates

    y = y₀+v₀y*t + 1/2 ay*t²

    where

    v₀y = initial velocity in the y direction = 0 m/s

    ay = acceleration in the x direction = - 0.5 m/s²

    y = coordinates in the y-axis

    y₀ = initial coordinate in the y-axis = 0

    then since y₀=0 and v₀y=0

    y = 1/2*ay*t²

    y = 1/2*ay*t² = 1/2 * (-0.5 m/s²) * (3 s) ² = - 2.25 m

    and

    vy=v₀y + ay*t = 0 + (-0.5 m/s²) * (3 s) = (-1.5 m/s)

    therefore the position vector (x, y) will be (1.5 m,-2.25 m)

    and the velocity vector (vx, vy) will be (0 m/s, - 1.5 m/s)
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