A particle moves along the x axis. It is initially at the position 0.350 m, moving with velocity 0.110 m/s and acceleration -0.380 m/s2. Suppose it moves with constant acceleration for 3.50 s.

(a) Find the position of the particle after this time.
(b) Find its velocity at the end of this time interval.
Next, assume it moves with simple harmonic motion for 3.50 s and x = 0 is its equilibrium position.
(c) Find its position.

(d) Find its velocity at the end of this time interval.

how do you know period is 3.5s

To solve these problems, we can use the equations of motion. The basic equations we will need are:

1. Position equation: x = x0 + v0t + (1/2)at^2
2. Velocity equation: v = v0 + at

Now let's solve these step-by-step:

(a) To find the position of the particle after 3.50s, we can use the position equation.

Given:
Initial position (x0) = 0.350 m
Initial velocity (v0) = 0.110 m/s
Acceleration (a) = -0.380 m/s^2
Time (t) = 3.50 s

Using the position equation, plug in the values:

x = x0 + v0t + (1/2)at^2
x = 0.350 + (0.110)(3.50) + (1/2)(-0.380)(3.50)^2

Calculating this expression will give us the position of the particle after 3.50s.

(b) To find the velocity of the particle at the end of 3.50s, we can use the velocity equation.

Given:
Initial velocity (v0) = 0.110 m/s
Acceleration (a) = -0.380 m/s^2
Time (t) = 3.50 s

Using the velocity equation, plug in the values:

v = v0 + at
v = 0.110 + (-0.380)(3.50)

Calculating this expression will give us the velocity of the particle at the end of 3.50s.

(c) Now, let's assume the particle moves with simple harmonic motion for 3.50s. We are given that x = 0 is its equilibrium position.

For simple harmonic motion, the equation of motion is given by:

x = A sin(ωt + φ)

Where:
x = position of the particle
A = amplitude of the motion
ω = angular frequency
t = time
φ = phase constant

Since the particle is at the equilibrium position when x = 0, we can conclude that φ = 0.

Given:
Time (t) = 3.50 s

Now we need to find the values of A (amplitude) and ω (angular frequency) in order to calculate the position.

(d) To find the velocity of the particle at the end of 3.50s in simple harmonic motion, we can differentiate the equation of motion with respect to time:

v = Aω cos(ωt + φ)

Since φ = 0, the equation becomes:

v = Aω cos(ωt)

Given:
Time (t) = 3.50 s

Now we need to substitute the values of A (amplitude) and ω (angular frequency) in order to calculate the velocity.

To find the answers to these questions, we will use the equations of motion under constant acceleration and the equations of simple harmonic motion.

(a) Find the position of the particle after 3.50 seconds:
To find the position of the particle, we can use the equation for position under constant acceleration:

x = x₀ + v₀t + (1/2)at²

Given:
x₀ (initial position) = 0.350 m
v₀ (initial velocity) = 0.110 m/s
a (acceleration) = -0.380 m/s²
t (time) = 3.50 s

Plugging in these values into the equation, we get:

x = 0.350 + (0.110)(3.50) + (1/2)(-0.380)(3.50)²

Solving this equation will give us the position of the particle after 3.50 seconds.

(b) Find its velocity at the end of this time interval:
To find the velocity of the particle, we can use the equation for velocity under constant acceleration:

v = v₀ + at

Given:
v₀ (initial velocity) = 0.110 m/s
a (acceleration) = -0.380 m/s²
t (time) = 3.50 s

Plugging in these values into the equation, we get:

v = 0.110 + (-0.380)(3.50)

Solving this equation will give us the velocity of the particle at the end of 3.50 seconds.

(c) Find its position when moving with simple harmonic motion for 3.50 seconds:
To find the position of the particle under simple harmonic motion, we can use the equation for position:

x = A * cos(ωt)

Given:
A (amplitude) = unknown
ω (angular frequency) = unknown
t (time) = 3.50 s

We need to know the values of A and ω to solve this equation. Without additional information, we cannot determine the particle's position under simple harmonic motion.

(d) Find its velocity at the end of this time interval when moving with simple harmonic motion:
To find the velocity of the particle under simple harmonic motion, we can use the equation for velocity:

v = -A * ω * sin(ωt)

Similarly to part (c), without additional information, we cannot determine the particle's velocity at the end of the 3.50-second interval under simple harmonic motion.

Set up the appropriate x(t) or V(t) equation and "plug in" the appropriate t. Thenk crank out the answer. to do, or at least attempt, yourself. Here are the equations:

(a) x = 0.11 t -0.19 t^2
(b) V = 0.11 - 0.38 t
(c) Since t = 3.5 s is the period, it returns to the starting point, x = 0, at that time. You don't need to know the amplitude
(d) Since t = 3.5 s is the period, it returns to the same starting velocity -0.11 m/s