Average and instantaneous velocity

Average velocity

• v_av = displacement ÷ elapsed time = [s(t₂) − s(t₁)]/(t₂ − t₁).
• Geometrically it is the slope of the chord joining the two points of the (position–time) curve.

Instantaneous velocity

• v = lim_Δt→0 [s(t + Δt) − s(t)]/Δt = ds/dt: velocity is the first derivative of position, i.e. the slope of the tangent to the (position–time) curve.
• The (scalar) speed is the magnitude |v| and is never negative. v > 0 means motion in the positive direction, v < 0 the opposite.
• The velocity vanishes at the instant the particle reverses its direction of motion.
• Average speed = total distance ÷ total time, and it is generally different from the magnitude of the average velocity.

Distance from the velocity curve

• Distance in [t₁, t₂] is the area between the (velocity–time) curve and the time axis: d = ∫_t₁^t₂ |v| dt.
• When the sign of v changes, split the interval at the instants where v = 0 and add the absolute values.

Worked example

1. A stone is thrown vertically upward so that its height after t seconds is s = 49t − 4.9t². Find the greatest height and the velocity when the height is 78.4 m.
   v = ds/dt = 49 − 9.8t. At the top v = 0 → t = 5 s, s_max = 245 − 122.5 = 122.5 m.
   49t − 4.9t² = 78.4 → divide by 4.9: t² − 10t + 16 = 0 → (t − 2)(t − 8) = 0 → t = 2 or 8.
   v(2) = 49 − 19.6 = +29.4 m/s (rising), v(8) = 49 − 78.4 = −29.4 m/s (falling); the speed is 29.4 m/s in both cases.