Particle Motion Analysis: Graphs & Max Position

Oct 13, 2025 by Dimemap Team 48 views

Hey guys! Let's dive into a fun physics problem involving a particle zipping along a straight line. We're given its velocity over time and need to figure out its acceleration and position, draw some cool graphs, and find out where it reaches its farthest point. Buckle up; it's gonna be an interesting ride!

Understanding the Problem

So, we've got this particle moving in one dimension, and we know its velocity (v) at different times (t). The initial position (x) at time t=0 is -14.6 meters. Our mission, should we choose to accept it, is threefold:

Create Acceleration-Time (a-t) and Position-Time (x-t) Graphs: We need to visualize how the particle's acceleration and position change over the 40 seconds.
Determine the Maximum Position: Find the largest x-coordinate the particle reaches during its journey.
Find the Time(s) of Maximum Position: Identify the specific time(s) when the particle is at that maximum position.

This problem combines kinematics (the study of motion) with graphical analysis. We'll use the relationships between position, velocity, and acceleration to build our graphs and solve for the unknowns. Remember those physics formulas? They're about to become our best friends!

Constructing the Acceleration-Time (a-t) Graph

Let's start with the acceleration-time graph. Remember that acceleration is the rate of change of velocity. Mathematically, it's the derivative of velocity with respect to time: a = dv/dt. Graphically, on a velocity-time graph, the acceleration is the slope of the line.

To create the a-t graph, we need to analyze the given v-t graph and determine the slope at different intervals. If the v-t graph is a straight line segment, the acceleration is constant during that interval and equal to the slope of the line. If the v-t graph is curved, the acceleration is changing, and we'd need calculus to find the instantaneous acceleration. However, for this problem, we'll assume the v-t graph consists of straight-line segments, making our lives much easier.

Example:

Interval 1 (0 < t < t1): If the velocity increases linearly from, say, 0 m/s to 10 m/s in 5 seconds, the acceleration is (10 m/s - 0 m/s) / (5 s) = 2 m/s². The a-t graph will show a horizontal line at a = 2 m/s² during this interval.
Interval 2 (t1 < t < t2): If the velocity remains constant at 10 m/s, the acceleration is 0 m/s² because the slope of the v-t graph is zero. The a-t graph will show a horizontal line at a = 0 m/s².
Interval 3 (t2 < t < t3): If the velocity decreases linearly from 10 m/s to 0 m/s in 3 seconds, the acceleration is (0 m/s - 10 m/s) / (3 s) = -3.33 m/s². The a-t graph will show a horizontal line at a = -3.33 m/s².

By repeating this process for each segment of the v-t graph, we can construct the complete a-t graph. Remember to pay attention to the sign of the slope – positive slope means positive acceleration (speeding up), and negative slope means negative acceleration (slowing down).

Important Considerations:

Units: Always include the correct units (m/s² for acceleration, m/s for velocity, s for time, and m for position).
Discontinuities: If the v-t graph has sharp corners (instantaneous changes in slope), the a-t graph will have discontinuities (jumps) at those points. In reality, instantaneous changes don't happen, but they're a useful approximation in many physics problems.

Constructing the Position-Time (x-t) Graph

Next up is the position-time graph. This one's a bit trickier, but we can handle it. Remember that velocity is the rate of change of position: v = dx/dt. This means the velocity is the slope of the x-t graph. To find the position at any given time, we need to integrate the velocity function with respect to time, or find the area under the v-t curve up to that time, and add it to the initial position.

x(t) = x(0) + ∫v(t) dt

Since we're likely dealing with piecewise linear functions for v(t), the integral becomes a sum of areas of rectangles and triangles. Let's break it down:

Example:

Interval 1 (0 < t < t1): If the velocity increases linearly from 0 m/s to 10 m/s, the position changes according to the equation x(t) = x(0) + area under the v-t curve. If the v-t graph is a straight line, the area is a triangle. So, x(t) = -14.6 m + (1/2) * t * (10 m/s). The x-t graph will be a curve (specifically, a parabola) during this interval.
Interval 2 (t1 < t < t2): If the velocity remains constant at 10 m/s, the position changes linearly: x(t) = x(t1) + 10 m/s * (t - t1). The x-t graph will be a straight line with a slope of 10 m/s.
Interval 3 (t2 < t < t3): If the velocity decreases linearly from 10 m/s to 0 m/s, the position changes according to x(t) = x(t2) + area under the v-t curve (another triangle or trapezoid). The x-t graph will be another parabolic curve.

By calculating the position at the end of each interval and plotting those points, we can sketch the x-t graph. The slope of the x-t graph at any point gives us the instantaneous velocity at that time.

Important Considerations:

Initial Position: Don't forget to add the initial position, x(0) = -14.6 m, to each position calculation. This shifts the entire x-t graph vertically.
Curvature: When the velocity is changing (non-zero acceleration), the x-t graph will be curved. When the velocity is constant, the x-t graph will be a straight line.
Smoothness: The x-t graph should be continuous, meaning there shouldn't be any sudden jumps in position. Even if the velocity changes abruptly, the position changes smoothly.

Determining the Maximum Position Coordinate

Now for the fun part: finding the maximum position. The maximum position occurs when the particle momentarily stops moving in the positive direction and starts moving in the negative direction, or at the end of the time interval if the velocity is always positive. In other words, the maximum position occurs when the velocity is zero (or at t=40 s).

To find the maximum position, we need to:

Identify Times When v(t) = 0: Look at the v-t graph and find the times where the velocity crosses the t-axis (i.e., v = 0).
Calculate Position at Those Times: Use the method described above (integrating the v-t graph or using kinematic equations) to calculate the position x(t) at each of those times.
Compare Positions: Compare the positions calculated in step 2 with the position at the end of the time interval (t = 40 s). The largest of these values is the maximum position coordinate.

Example:

Let's say the velocity is zero at t = 15 s and t = 30 s. We would calculate x(15 s), x(30 s), and x(40 s). The largest of these three values would be the maximum position.

Important Considerations:

Multiple Maxima: It's possible for the particle to have multiple local maxima (peaks) in its position. Make sure to check all of them to find the absolute maximum.
End Points: Don't forget to check the position at the endpoints of the time interval (t = 0 and t = 40 s) as the maximum position could occur there.

Determining the Time(s) of Maximum Position

This part is easy! Once you've found the maximum position, the time(s) at which it occurs are simply the values of t you used to calculate that maximum position. In the example above, if x(30 s) was the maximum position, then the time of maximum position would be t = 30 s.

Putting It All Together

To recap, here's the process:

Analyze the v-t graph: Determine the slopes (accelerations) and areas under the curve (changes in position).
Construct the a-t graph: Plot the acceleration as a function of time.
Construct the x-t graph: Calculate and plot the position as a function of time, remembering the initial position.
Find the times when v(t) = 0: These are potential times of maximum position.
Calculate the position at those times and at the endpoints (t=0 and t=40):
Identify the maximum position and the corresponding time(s).

By carefully following these steps, you can successfully analyze the motion of the particle, create the required graphs, and determine the maximum position and the time(s) at which it occurs. Good luck, and happy graphing!