A particle starts from the origin with a velocity of 6.0 i m/s and moves in the xy plane with a constant acceleration of (-2.0i + 4.0j)m/s^2 . at he instant the particle achieves maximum positive x-coordinate, how far is it from the origin?
To find the distance from the origin when the particle achieves maximum positive x-coordinate, we need to determine the time it takes for the particle to reach that point.
Given:
Initial velocity (π’) = 6.0 π m/s
Acceleration (π) = -2.0 π + 4.0 π m/sΒ²
At maximum positive x-coordinate, the y-component of velocity becomes zero (π£π¦ = 0).
Let's find the time it takes for π£π¦ to become zero.
Using the formula: π£π¦ = π’π¦ + ππ¦π‘
Since π£π¦ = 0,
0 = 6.0 + 4.0π‘
Solving the above equation for time (π‘):
4.0π‘ = -6.0
π‘ = -6.0/4.0
π‘ = -1.5 seconds
The time it takes for the particle to reach maximum positive x-coordinate is 1.5 seconds.
To find the distance from the origin, we can use either the displacement formula or the equation of motion.
Using the equation of motion: π = π’π‘ + 0.5ππ‘Β²
For x-coordinate:
π π₯ = (6.0π)(-1.5) + 0.5(-2.0π)(-1.5)Β²
π π₯ = -9.0π + 0.5(4.5)π
π π₯ = -9.0π + 2.25π
π π₯ = -6.75π
For y-coordinate:
π π¦ = (0.0π)(-1.5) + 0.5(4.0π)(-1.5)Β²
π π¦ = 0.0π + 0.5(9.0)π
π π¦ = 0.0π + 4.5π
π π¦ = 4.5π
The displacement from the origin when the particle achieves maximum positive x-coordinate is given by:
π = π π₯ + π π¦
π = -6.75π + 4.5π
The magnitude of the displacement is:
|π | = β((-6.75)Β² + (4.5)Β²) (using Pythagoras theorem)
|π | = β(45.5625 + 20.25)
|π | = β65.8125
|π | β 8.12
Therefore, the particle is approximately 8.12 units away from the origin when it achieves maximum positive x-coordinate.
To find the distance of the particle from the origin when it achieves its maximum positive x-coordinate, we need to determine the time it takes for the particle to reach that point.
Starting from the origin, we know the initial velocity of the particle is 6.0 i m/s, and the acceleration is (-2.0 i + 4.0 j) m/s^2.
We can use the kinematic equation to determine the time it takes for the particle to reach the maximum positive x-coordinate:
x = xβ + vβt + 0.5atΒ²
Since the velocity is constant in the y-direction (j-direction), the y-coordinate remains zero throughout the motion. Therefore, the y-component of the equation is not required.
For the x-coordinate, we have:
x = 0 + (6.0 i) t + 0.5 (-2.0 i) tΒ²
Simplifying the equation gives us:
x = 6.0t - tΒ²
To determine the time when the particle achieves its maximum positive x-coordinate, we take the derivative of the equation with respect to t and set it to zero:
dx/dt = 6.0 - 2t = 0
Solving for t gives us:
t = 3.0 s
Now that we have the time at which the particle achieves its maximum positive x-coordinate, we can substitute it back into the equation to find the distance from the origin:
x = 6.0(3.0) - (3.0)Β²
x = 18.0 - 9.0
x = 9.0 meters
Therefore, the particle is 9.0 meters away from the origin when it achieves its maximum positive x-coordinate.