A particle leaves the origin with an initial velocity vector v = (3.00 I ) m/s a

Question

A particle leaves the origin with an initial velocity vector v = (3.00 I ) m/s and a constant acceleration vector a = (-1.00 I -0.650 j ). What is the velocity of the particle when it reaches its maximum x coordinate ? i- component of velocity ? Tries 0/10 j- component of velocity ? Tries 0/10 When does it reach its maximum x coordinate ? Tries 0/10 When does it reach its maximum x coordinates ? Tries 0/10 Where is the particle at this time ? i- component of position? Tries 0/10 j- component of position ? Tries 0/10

Explanation / Answer

Vx = Vxo + at

we can find the time it takes for the x component of speed to reach zero (and thus for the particle to reach its maximum x coordinate).

vx =0

0 = 3 + (-1)t
t = 3 / 1
t = 3seconds

The velocity at this point will be entirely in the y-direction, and can be calculated as:

Vy = Vyo + at
Vy = 0 -0.65(3)
Vy = -1.95 m/s

The position can be calculated separately for the x and y components using the equation:

X = Xo + VoT + 1/2 aT^2
X = 0 + 3(3) + 1/2 (-1)(3)^2
X = 4.5

(Similarly for Y)

Y = 0 + 0 + 1/2 (-0.65)(3)^2
Y = - 2.925

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