Math Problem Statement

At t = 2.0 s, what is the particle's acceleration? Express your answer with the appropriate units.

Solution

The graph you provided displays the velocity vx(t)v_x(t) versus time tt for a particle, and the question asks for the particle's acceleration at t=2.0st = 2.0 \, \text{s}.

To determine acceleration:

  • Acceleration is the rate of change of velocity with respect to time, defined as a=ΔvΔta = \frac{\Delta v}{\Delta t}.
  • In this case, the graph shows a linear relationship, meaning that the slope of the velocity-time graph represents the constant acceleration.

Step-by-step solution:

  1. Identify points from the graph:

    • At t=0st = 0 \, \text{s}, the velocity vx=6m/sv_x = 6 \, \text{m/s}.
    • At t=3st = 3 \, \text{s}, the velocity vx=0m/sv_x = 0 \, \text{m/s}.
  2. Calculate the slope of the line (acceleration): The slope of the line is: a=ΔvxΔt=vx2vx1t2t1a = \frac{\Delta v_x}{\Delta t} = \frac{v_{x2} - v_{x1}}{t_2 - t_1} Using the points (t1,vx1)=(0s,6m/s)(t_1, v_{x1}) = (0 \, \text{s}, 6 \, \text{m/s}) and (t2,vx2)=(3s,0m/s)(t_2, v_{x2}) = (3 \, \text{s}, 0 \, \text{m/s}): a=0m/s6m/s3s0s=6m/s3s=2m/s2a = \frac{0 \, \text{m/s} - 6 \, \text{m/s}}{3 \, \text{s} - 0 \, \text{s}} = \frac{-6 \, \text{m/s}}{3 \, \text{s}} = -2 \, \text{m/s}^2

Result:

The acceleration is constant throughout the time interval, so at t=2.0st = 2.0 \, \text{s}, the particle's acceleration is: a=2m/s2a = -2 \, \text{m/s}^2

Would you like further details or have any questions? Here are some related questions you could explore:

  1. What would be the particle's velocity at t=1st = 1 \, \text{s}?
  2. How far has the particle traveled by t=2.0st = 2.0 \, \text{s}?
  3. What is the particle's displacement during the entire 3-second interval?
  4. What would happen to the motion if the acceleration was positive instead of negative?
  5. Can we calculate the average velocity of the particle between t=0st = 0 \, \text{s} and t=2.0st = 2.0 \, \text{s}?

Tip: In velocity-time graphs, the slope gives you acceleration, and the area under the curve gives you displacement.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Velocity-Time Graphs
Acceleration

Formulas

a = Δv / Δt

Theorems

Slope of Velocity-Time Graph

Suitable Grade Level

Grades 9-12