Average Velocity and acceleration

Question

Initially, a cyclist moves forward with a speed of 16.0 m/s. After advancing 200 m while undergoing non-constant acceleration, the cyclist moves forward with a speed of 4.00 m/s. If the time between the start and end of the 200 m trip equals 16.0 s, what are the magnitudes and the directions of the average velocity, vavg, and the average acceleration, aavg, of the cyclist?

Answer 1

Let

v₁ = 16 m/s be the initial speed,

t₁ (unknown) be the time when the initial speed is measured,

v₁ = 4 m/s be the final speed,

t₂ (unknown) be the time when the final speed is measured,

Δt = t₂ - t₁ = 16 s be the time interval,

Δs = 200 m be the distance travelled.

vavg = Δs/Δt = 200m/16s = 12.5m/s

The average velocity is positive, i.e., in the forward direction.

It is positive because Δs is positive, i.e., she advanced forward 200m.

aavg = (v₂-v₁)/(t₂-t₁) = (-12m/s)/16s = -0.75m/s²

The average acceleration is negative, i.e., it an an acceleration in be backward direction, or

deceleration in the forward direction.