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?

Question

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?

Answer 1

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 (𝑢) = 6.0 𝑖 m/s
Acceleration (𝑎) = -2.0 𝑖 + 4.0 𝑗 m/s²

At maximum positive x-coordinate, the y-component of velocity becomes zero (𝑣𝑦 = 0).

Let's find the time it takes for 𝑣𝑦 to become zero.

Using the formula: 𝑣𝑦 = 𝑢𝑦 + 𝑎𝑦𝑡

Since 𝑣𝑦 = 0,
0 = 6.0 + 4.0𝑡

Solving the above equation for time (𝑡):
4.0𝑡 = -6.0
𝑡 = -6.0/4.0
𝑡 = -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: 𝑠 = 𝑢𝑡 + 0.5𝑎𝑡²

For x-coordinate:
𝑠𝑥 = (6.0𝑖)(-1.5) + 0.5(-2.0𝑖)(-1.5)²
𝑠𝑥 = -9.0𝑖 + 0.5(4.5)𝑖
𝑠𝑥 = -9.0𝑖 + 2.25𝑖
𝑠𝑥 = -6.75𝑖

For y-coordinate:
𝑠𝑦 = (0.0𝑗)(-1.5) + 0.5(4.0𝑗)(-1.5)²
𝑠𝑦 = 0.0𝑗 + 0.5(9.0)𝑗
𝑠𝑦 = 0.0𝑗 + 4.5𝑗
𝑠𝑦 = 4.5𝑗

The displacement from the origin when the particle achieves maximum positive x-coordinate is given by:
𝑠 = 𝑠𝑥 + 𝑠𝑦
𝑠 = -6.75𝑖 + 4.5𝑗

The magnitude of the displacement is:
|𝑠| = √((-6.75)² + (4.5)²) (using Pythagoras theorem)
|𝑠| = √(45.5625 + 20.25)
|𝑠| = √65.8125
|𝑠| ≈ 8.12

Therefore, the particle is approximately 8.12 units away from the origin when it achieves maximum positive x-coordinate.

Answer 2

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 = x₀ + v₀t + 0.5at²

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) t²

Simplifying the equation gives us:
x = 6.0t - t²

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)²
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.

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.

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.

20m