At t = 0,a particle leaves the origin with a velocity of 12 m/s in the positive x direction and moves in the xy plane with a constant acceleration of $( - 2.0 \tilde { \mathbf { i } } + 4.0 \tilde { \mathbf { j } } ) \mathrm { m } / \mathrm { s } ^ { 2 }$ .At the instant the y coordinate of the particle is 18 m,what is the x coordinate of the particle?
A)30 m
B)21 m
C)27 m
D)24 m
E)45 m

Question

Quizplus · Accepted Answer

We can begin by using the kinematic equation:
y = yo + voy*t + 1/2*ay*t^2

where y=18m, yo=0, voy=0 (since the velocity is entirely in the x direction), and ay=11ea99f6_d755_562c_8bb6_71fa4c8757ac_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11.

Solving for t, we get:
t = sqrt(2y/ay) = sqrt(2*18/11ea99f6_d755_562c_8bb6_71fa4c8757ac_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11_TB5764_11) = 2 sec (approx)

Now we can use the kinematic equation for x:
x = xo + vox*t + 1/2*ax*t^2

where xo=0, vox=12 m/s, and ax=0 (since the acceleration is entirely in the y direction).

Plugging in t=2 sec, we get:
x = 0 + 12*2 + 1/2*0*(2)^2 = 24 m

Therefore, the x coordinate of the particle is 24 m, which corresponds to choice C.

At T = 0,a Particle Leaves the Origin with a Velocity