Find the position and velocity of a particle at t = 1.92 s if the particle is initially moving east at a speed of 18.4 m/s and experiences an acceleration of magnitude 3.96 m/s2, directed west.
To find the position and velocity of a particle at a specific time, we need to use kinematic equations. In this case, we have the initial velocity, acceleration, and time, and we need to find the position and velocity.
Let's break down the given information:
Initial velocity (u) = 18.4 m/s (east)
Acceleration (a) = -3.96 m/s² (westward)
Time (t) = 1.92 s
First, let's find the final velocity (v) of the particle using the equation:
v = u + at
Substituting the known values:
v = 18.4 m/s - 3.96 m/s² * 1.92 s
v = 18.4 m/s - 7.6032 m/s
v = 10.7968 m/s (westward)
So, the velocity of the particle at t = 1.92 s is 10.7968 m/s in the westward direction.
Next, let's find the displacement (s) of the particle using the equation:
s = ut + (1/2)at²
Substituting the known values:
s = 18.4 m/s * 1.92 s + (1/2)(-3.96 m/s²)(1.92 s)²
s = 35.328 m + (-7.6032 m/s²)(3.6864 s²)
s = 35.328 m + (-27.986 m)
s = 7.342 m
Therefore, the position of the particle at t = 1.92 s is a displacement of 7.342 m (east).
To summarize:
- Velocity at t = 1.92 s: 10.7968 m/s (westward).
- Position at t = 1.92 s: 7.342 m (east).