A computer model displays the motion of a particle on a coordinate system in real time. At time t = 0, the particle is at the origin of the coordinate system and has velocity components vx = 0 and vy = 6.8 m/s. The particle has acceleration components of 
ax = −3.2 m/s2
 and ay = 0.
(a)
What are the x and y positions of the particle, in meters, at t = 4.0 s?
x	=	
y	=	
(b)
What are velocity components of the particle, in m/s, at t = 4.0 s?
vx	=	
vy	=	
(c)
How does the speed of the particle change from t = 0 to t = 4.0 s?

Let's break down the problem into its parts. We know the particle's velocity, acceleration components, and initial conditions, and we need to calculate the position and velocity at \( t = 4.0 \, \text{s} \).

### Given Data:
- Initial position: \( x_0 = 0 \, \text{m}, \, y_0 = 0 \, \text{m} \)
- Initial velocities: \( v_{x0} = 0 \, \text{m/s}, \, v_{y0} = 6.8 \, \text{m/s} \)
- Accelerations: \( a_x = -3.2 \, \text{m/s}^2, \, a_y = 0 \, \text{m/s}^2 \)
- Time: \( t = 4.0 \, \text{s} \)

We will solve the problem using the kinematic equations of motion.

### Part (a): Find the \( x \)- and \( y \)-positions of the particle at \( t = 4.0 \, \text{s} \).

The general equations for position under constant acceleration are:

\[
x = x_0 + v_{x0} t + \frac{1}{2} a_x t^2
\]
\[
y = y_0 + v_{y0} t + \frac{1}{2} a_y t^2
\]

#### x-position:
\[
x = 0 + (0)(4.0) + \frac{1}{2} (-3.2)(4.0)^2
\]
\[
x = \frac{1}{2} (-3.2)(16) = -25.6 \, \text{m}
\]

#### y-position:
Since \( a_y = 0 \), the y-component of motion is uniform, so:
\[
y = 0 + (6.8)(4.0) + \frac{1}{2} (0)(4.0)^2
\]
\[
y = 27.2 \, \text{m}
\]

So, at \( t = 4.0 \, \text{s} \):
\[
x = -25.6 \, \text{m}, \quad y = 27.2 \, \text{m}
\]

### Part (b): Find the velocity components at \( t = 4.0 \, \text{s} \).

The velocity equations under constant acceleration are:

\[
v_x = v_{x0} + a_x t
\]
\[
v_y = v_{y0} + a_y t
\]

#### x-component of velocity:
\[
v_x = 0 + (-3.2)(4.0) = -12.8 \, \text{m/s}
\]

#### y-component of velocity:
Since \( a_y = 0 \), the y-component remains constant:
\[
v_y = 6.8 + (0)(4.0) = 6.8 \, \text{m/s}
\]

So, the velocity components at \( t = 4.0 \, \text{s} \) are:
\[
v_x = -12.8 \, \text{m/s}, \quad v_y = 6.8 \, \text{m/s}
\]

### Part (c): How does the speed of the particle change from \( t = 0 \) to \( t = 4.0 \, \text{s} \)?

The speed of the particle is the magnitude of the velocity vector:

At \( t = 0 \):
\[
v(0) = \sqrt{v_{x0}^2 + v_{y0}^2} = \sqrt{0^2 + 6.8^2} = 6.8 \, \text{m/s}
\]

At \( t = 4.0 \):
\[
v(4.0) = \sqrt{v_x^2 + v_y^2} = \sqrt{(-12.8)^2 + (6.8)^2}
\]
\[
v(4.0) = \sqrt{163.84 + 46.24} = \sqrt{210.08} \approx 14.49 \, \text{m/s}
\]

Thus, the speed of the particle increases from \( 6.8 \, \text{m/s} \) at \( t = 0 \) to approximately \( 14.49 \, \text{m/s} \) at \( t = 4.0 \, \text{s} \).

### Summary:
- (a) \( x = -25.6 \, \text{m}, \quad y = 27.2 \, \text{m} \)
- (b) \( v_x = -12.8 \, \text{m/s}, \quad v_y = 6.8 \, \text{m/s} \)
- (c) Speed increases from \( 6.8 \, \text{m/s} \) to approximately \( 14.49 \, \text{m/s} \).

Would you like more details or further clarifications?

Here are five related questions:
1. How would the results change if the acceleration in the y-direction were non-zero?
2. What would happen to the particle's path if the x-component of velocity were initially non-zero?
3. How can we determine the total displacement of the particle after 4.0 seconds?
4. What is the angle of the velocity vector at \( t = 4.0 \, \text{s} \)?
5. How would the motion differ if the accelerations were reversed in sign?

**Tip:** When calculating motion under constant acceleration, always check if any components of velocity or acceleration are zero to simplify the equations.

A computer model displays the motion of a particle on a coordinate system in real time. At time t = 0, the particle is at the origin of the coordinate system and has velocity components vx = 0 and vy = 6.8 m/s. The particle has acceleration components of ax = −3.2 m/s2 and ay = 0.
(a) What are the x and y positions of the particle, in meters, at t = 4.0 s?
(b) What are velocity components of the particle, in m/s, at t = 4.0 s?
(c) How does the speed of the particle change from t = 0 to t = 4.0 s?

Calculate the position and velocity of a particle in 2D motion under constant acceleration. At t = 4.0s, determine the particle's x and y positions, velocity components, and how its speed changes over time. Suitable for high school students (Grades 10-12) studying kinematics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation