Position Physics Example 2

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Example 2

medium
A particle's position is given by x(t)=4t2−2t+1x(t) = 4t^2 - 2t + 1 (in meters, with tt in seconds). What is the particle's position at t=0 st = 0 \text{ s} and t=3 st = 3 \text{ s}? What is the displacement between these times?

Solution

  1. 1
    At t=0t = 0: x(0)=4(0)2−2(0)+1=1 mx(0) = 4(0)^2 - 2(0) + 1 = 1 \text{ m}.
  2. 2
    At t=3t = 3: x(3)=4(9)−2(3)+1=36−6+1=31 mx(3) = 4(9) - 2(3) + 1 = 36 - 6 + 1 = 31 \text{ m}.
  3. 3
    Displacement: Δx=x(3)−x(0)=31−1=30 m\Delta x = x(3) - x(0) = 31 - 1 = 30 \text{ m}

Answer

x(0)=1 m,x(3)=31 m,Δx=30 mx(0) = 1 \text{ m}, \quad x(3) = 31 \text{ m}, \quad \Delta x = 30 \text{ m}
A position function gives the location of an object at any time. By evaluating the function at two different times and subtracting, we find the displacement over that interval.

About Position

The location of an object relative to a chosen reference point (origin), described using coordinates in a given reference frame.

Learn more about Position →

More Position Examples

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Example 3 medium

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Example 4 hard

A particle's position is [formula] (meters). At what times is the particle at position [formula]? Wh

← All Position Examples Position Concept

Related Concepts

DisplacementVelocity