Answer:
Option C : 60 meters
Step-by-step explanation:
It is given that a particle starts from the origin at t = 0 with an initial velocity of 3.0i m/s and moves in the x-y plane with a constant acceleration of (6.0i + 4.0j) m/s².
For clarity, in this context, 'i' corresponds to the x-direction, while 'j' corresponds to the y-direction.
Since the acceleration is constant, we can use one of the kinematic equations of motion (SUVAT equations) to find the x-coordinate of the particle when its y-coordinate is 32 meters:
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
where:
- s = displacement in m
- u = initial velocity in m/s
- v = final velocity in m/s
- a = acceleration in m/s²
- t = time in seconds
Substitute the given i and j vectors for each parameter into the displacement equation:
[tex](x\mathbf{i}+32\mathbf{j})=(3.0\mathbf{i}+0.0\mathbf{j})t+\dfrac{1}{2}(6.0\mathbf{i} + 4.0\mathbf{j})t^2[/tex]
Resolve horizontally by focussing only on the i components of each vector:
[tex]x\mathbf{i}=3.0\mathbf{i}t+\dfrac{1}{2}(6.0\mathbf{i})t^2\\\\\\x=3.0t+\dfrac{1}{2}(6)t^2\\\\\\x=3t+3t^2[/tex]
Resolve vertically by focussing only on the j components of each vector:
[tex]32\mathbf{j}=0.0\mathbf{j}t+\dfrac{1}{2}(4.0\mathbf{j})t^2\\\\\\32=(0)t+\dfrac{1}{2}(4)t^2\\\\\\32=2t^2[/tex]
Solve for t:
[tex]32=2t^2\\\\t^2=16\\\\t=4[/tex]
Substitute the found value of t into x = 3t + 3t²:
[tex]x=3(4)+3(4)^2\\\\x=12+3(16)\\\\x=12+48\\\\x=60[/tex]
Therefore, the x-coordinate of the particle at the instant when its y-coordinate is 32 m is 60 meters. Therefore:
[tex]\LARGE\boxed{\boxed{D=60}}[/tex]
