Math Problem Statement

The equation gives the position s = f(t) of a body moving on a coordinate line (s in meters, t in seconds).

s (m)

What is the body's velocity when t = 2.5 sec?

What is the body's velocity when t = 2.5 sec?

Solution

The image shows a graph of position s=f(t)s = f(t) (in meters) as a function of time tt (in seconds). To find the velocity of the body at t=2.5t = 2.5 seconds, we need to determine the slope of the position function at that specific time. The velocity is the rate of change of position with respect to time, which is the derivative of the position function f(t)f(t).

Step-by-step process:

  1. Locate t=2.5t = 2.5 on the graph: From the graph, t=2.5t = 2.5 lies between t=2t = 2 and t=3t = 3. We need to find the slope of the graph in this interval.

  2. Analyze the slope in the interval 2t32 \leq t \leq 3:

    • The graph shows a flat line between t=2t = 2 and t=3t = 3. This indicates that the position of the body does not change, meaning that the velocity is 00 during this time interval.
  3. Conclusion: Since the slope of the position graph is zero at t=2.5t = 2.5, the velocity of the body at t=2.5t = 2.5 seconds is 0\boxed{0} meters per second.

Let me know if you'd like further details or explanations!

Here are five related questions for deeper exploration:

  1. What is the body's acceleration at t=2.5t = 2.5 seconds?
  2. How would you calculate the velocity at t=4t = 4 seconds?
  3. What is the body's total displacement from t=0t = 0 to t=5t = 5?
  4. How would the velocity change if the slope of the graph between t=2t = 2 and t=3t = 3 were steeper?
  5. How can you use the graph to find when the body changes direction?

Tip: Always check if the graph is increasing, decreasing, or flat to determine the sign and magnitude of velocity.

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Math Problem Analysis

Mathematical Concepts

Position Function
Velocity
Rate of Change
Slope of a Function

Formulas

Velocity = Derivative of Position Function (ds/dt)
Slope = Δy/Δx

Theorems

Derivative as Rate of Change
Graphical Interpretation of Derivatives

Suitable Grade Level

Grades 9-12