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A particle's (x, t) coordinates at two instants are A (x=-14m, t=3s) and B (x=6m, t=1s). What is the particle's average velocity during this time?

Question

A particle's (x, t) coordinates at two instants are A (x=-14m, t=3s) and B (x=6m, t=1s). What is the particle's average velocity during this time?

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Solution

The average velocity of a particle is calculated by the change in displacement over the change in time.

In this case, the displacement (Δx) is the final position minus the initial position. So, Δx = x(final) - x(initial) = 6m - (-14m) = 20m.

The change in time (Δt) is the final time minus the initial time. So, Δt = t(final) - t(initial) = 1s - 3s = -2s.

Therefore, the average velocity (v) is Δx / Δt = 20m / -2s = -10 m/s.

So, the particle's average velocity during this time is -10 m/s. The negative sign indicates that the particle is moving in the opposite direction to the positive x direction.

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