Investigating Particle's Acceleration from Position-Time Graph

[lauriecherie]

Homework Statement

The figure below depicts the motion of a particle moving along an x-axis with a constant acceleration. What are the magnitude and direction of the particle's acceleration?

Can anyone explain to me where to begin on this one? It's a graph of position of time. It's concave up going through points (0,-2), (1,0), and (2,6).

The answer should be in m/s^2. It also let's me choose between -x, +x, -y, and +y for direction.

Homework Equations

The Attempt at a Solution

We didn't do any examples of this in class. I'm not sure where to begin on this one. Of course I don't expect an exact answer, but can someone walk me through it with an explanation. I know x(t)'= v(t) and x(t)"= v(t)'= a(t).

 

[Nabeshin]

One thing you could try is to fit a curve to these three data points and then use the differentiation formulas you wrote down to find the acceleration of the particle.

 

[nlightner]

To begin, let's first understand what the graph is showing us. The x-axis represents time, and the y-axis represents position. The fact that the graph is concave up means that the particle is accelerating. The steeper the slope of the graph, the greater the acceleration.

To find the magnitude of the acceleration, we can use the formula a = Δv/Δt, where Δv is the change in velocity and Δt is the change in time. We can find the change in velocity by looking at the slope of the graph. At point (0,-2), the slope of the graph is 2 (rise over run). This means that the velocity is increasing by 2 m/s every second. Similarly, at point (1,0), the slope is 6, meaning the velocity is increasing by 6 m/s every second. Therefore, the change in velocity is 6-2 = 4 m/s. The change in time is 1 second, since the particle moves from (0,-2) to (1,0) in 1 second.

Plugging these values into the formula, we get a = 4/1 = 4 m/s^2. Therefore, the magnitude of the acceleration is 4 m/s^2.

To determine the direction of the acceleration, we can use the sign of the slope. Since the slope is positive, the acceleration is in the positive direction, or +x. The particle is moving towards larger values on the x-axis.

In summary, the particle's acceleration is 4 m/s^2 in the +x direction.