Find the position and velocity of a particle at t=1.94 s if the particle is initially moving east at a speed of 19.8 m/s and experiences an acceleration of magnitude 4.08 m/s^2, directed west.

Magnitude and direction of the position.
magnitude-
direction- East of its original position

Magnitude and direction of the velocity.
magnitude-
direction- towards the east

I have figured out the directions for both of them, but I have been unable to figure out the magnitudes, I have tried many equations but none seem to work correctly. An answer with a detailed explanation is greatly appreciated.

Chris

V = 19.8m/s, East.

V = at=4.08m/s^2*1.94s = 7.92m/s,West.

Vn = 19.8 - 7.92 = 11.88m/s, East.
Vn = Net velocity.

To find the position and velocity of the particle at t=1.94s, we can use the equations of motion. Let's break it down step by step.

Step 1: Find the position of the particle at t=1.94s.
The initial velocity of the particle is given as 19.8 m/s towards the east. We need to find the displacement of the particle at t=1.94s.

To do this, we can use the equation:
displacement = initial velocity * time + 0.5 * acceleration * time^2

Here, the initial velocity is 19.8 m/s towards the east, the time is 1.94s, and the acceleration is 4.08 m/s^2 towards the west.

Let's plug in the values:
displacement = 19.8 m/s * 1.94 s + 0.5 * (-4.08 m/s^2) * (1.94 s)^2

Calculating the right side of the equation:
displacement = 19.8 m/s * 1.94 s - 0.5 * 4.08 m/s^2 * 3.7636 s^2
displacement = 38.412 m - 7.6536 m
displacement = 30.7584 m

The displacement is 30.7584 meters towards the east of the particle's original position.

Step 2: Find the velocity of the particle at t=1.94s.
The velocity of the particle at t=1.94s is the sum of the initial velocity and the acceleration, both corrected for the time elapsed.

To do this, we can use the equation:
velocity = initial velocity + acceleration * time

Using the given values:
velocity = 19.8 m/s + (-4.08 m/s^2) * 1.94 s

Calculating the right side of the equation:
velocity = 19.8 m/s - 7.9152 m/s
velocity = 11.8848 m/s

The magnitude of the velocity is 11.8848 m/s towards the east.

So, the position of the particle at t=1.94s is 30.7584 meters towards the east of its original position, and its velocity at that time is 11.8848 m/s towards the east.

To find the position and velocity of a particle at a specific time, we can use kinematic equations. Let's break down the problem step by step.

Step 1: Acceleration and initial velocity
Given:
Initial velocity (u) = 19.8 m/s (east)
Acceleration (a) = 4.08 m/s^2 (west)

Step 2: Finding the final velocity (v)
We can use the equation:
v = u + at

Substituting the values:
v = 19.8 m/s + (4.08 m/s^2)(1.94 s)
v = 19.8 m/s + 7.9152 m/s
v = 27.7152 m/s

So, the magnitude of the final velocity (speed) is 27.7152 m/s, and the direction is towards the east.

Step 3: Finding the position (displacement)
We can use the equation:
s = ut + (1/2)at^2

Substituting the values:
s = (19.8 m/s)(1.94 s) + (1/2)(4.08 m/s^2)(1.94 s)^2
s = 38.412 m + (1/2)(4.08 m/s^2)(3.7636 s^2)
s = 38.412 m + (0.5)(4.08 m/s^2)(14.1802096 s^2)
s = 38.412 m + 28.99217392 m
s = 67.40417392 m

So, the magnitude of the position (displacement) is 67.40417392 m, and the direction is east of its original position.

To summarize:
Magnitude and direction of the position:
- Magnitude: 67.40417392 m
- Direction: East of its original position

Magnitude and direction of the velocity:
- Magnitude: 27.7152 m/s
- Direction: Towards the east