Physics College

Find the position and velocity of a particle at t = 1.98 s if the particle is initially moving east at a speed of 20.4 m/s and experiences an acceleration of magnitude 4.40 m/s2, directed west.

Magnitude and direction of the position.
magnitude ---blank---
direction ---Select---

Magnitude and direction of the velocity.
magnitude ---blank---
direction ---select---

Answers

Answer 1
The position of the particle at t = 1.98 s is 31.7006 meters east of its initial position.The velocity of the particle at t = 1.98 s is 11.688 m/s to the east.

Position and velocity of particles

To find the position and velocity of a particle at a specific time, we can use the equations of motion.

Given:

Initial velocity (u) = 20.4 m/s (east)

Acceleration (a) = -4.40 m/s² (west)

Time (t) = 1.98 s

To find the position (displacement) of the particle at time t, we can use the equation:

s = ut + (1/2)at²

s = (20.4 m/s)(1.98 s) + (1/2)(-4.40 m/s²)(1.98 s)²

s = 40.392 m + (1/2)(-4.40 m/s²)(3.9204 s²)

s = 40.392 m - 8.6914 m

s ≈ 31.7006 m

To find the velocity of the particle at time t, we can use the equation:

v = u + at

v = (20.4 m/s) + (-4.40 m/s²)(1.98 s)

v = 20.4 m/s - 8.712 m/s

v ≈ 11.688 m/s

Therefore, the velocity of the particle at t = 1.98 s is approximately 11.688 m/s to the east.

More on displacement and velocity can be found here: https://brainly.com/question/28804440

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