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Define kinematic quantities; a = −2 m/s² distance

Short answer

Position is location, velocity is rate of change of position, acceleration is rate of change of velocity. With a = -2 m/s², v₀=10 m/s, at t=5 s: v=0 m/s, position 25 m, total distance 25 m (motion in one direction).

Define position, velocity and acceleration of a particle. The acceleration of a particle is defined by the relation a = −2 m/s². Initially if the velocity of the particle is 10 m/s at 0 m position, determine the velocity, position and the total distance travelled at the instant of 5 seconds.

Definitions

  • Position (x): the location of the particle on its path, measured from a chosen origin (m).
  • Velocity (v): the rate of change of position, v = dx/dt (m/s). Average velocity is Δx/Δt.
  • Acceleration (a): the rate of change of velocity, a = dv/dt = v dv/dx (m/s²).

Problem

a = −2 m/s², v₀ = 10 m/s, x₀ = 0.

  • v = v₀ + at = 10 − 2t
  • x = 10t − t²

At t = 5 s:

  • v = 10 − 2(5) = 0 m/s
  • x = 10(5) − 25 = 25 m

Total distance: the velocity is zero at t = 10/2 = 5 s and positive before that, so the particle moves in one direction only up to t = 5 s. The distance travelled equals the displacement:

Total distance = 25 m. (If the motion is continued beyond 5 s, the particle returns; at 5 s it has just stopped.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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