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Lesson 6 of 10

Average Speed vs Average Velocity vs Instantaneous Velocity

平均速度、平均速率、瞬时速度

Today's question

Two taxi rides to the same place: 7.5 km in 14 min, or 10 km in 15 min. Which one is faster?

Lesson notes

A short written version of Lesson 6. The study sheet has more practice questions.

Short answer

Trip 1 takes less time for the same trip, so its average velocity is greater. Trip 2 covers 0.67 km/min versus 0.54 km/min, so its average speed is greater.

Key ideas

  • "Fast" has two meanings. Covering ground fast is distance ÷ time. Getting there fast is displacement ÷ time.
  • Average velocity = displacement ÷ time, v = x / t, in m/s. It is a vector. Its direction is the direction of the displacement.
  • Average velocity only looks at the start and the end, not the path. It belongs to a time interval, not an instant. Always give its direction.
  • Average speed = distance ÷ time, in m/s. It is a scalar. Average speed is NOT the magnitude of average velocity.
  • Instantaneous velocity is the average velocity over an extremely short time, such as 0.0001 s. Its direction is along the tangent to the path. Instantaneous speed is its magnitude.
  • Average speed ≥ magnitude of average velocity, because the straight line is the shortest path. They are equal only in one-way straight-line motion.

Step by step

  1. Both trips start at home and end at the same place, so both trips have the same displacement.
  2. Trip 1 takes 14 min and Trip 2 takes 15 min, so Trip 1 gets there sooner.
  3. Same displacement in less time means Trip 1 has the greater average velocity (displacement ÷ time).
  4. Average speed uses distance ÷ time. Trip 1: 7.5 km ÷ 14 min ≈ 0.54 km/min.
  5. Trip 2: 10 km ÷ 15 min ≈ 0.67 km/min. Trip 2 covers more ground per minute, so it has the greater average speed.
  6. So Trip 1 is faster at getting there, and Trip 2 is faster at covering ground.

Common mistakes

  • Wrong: treating average speed as the magnitude of average velocity. Right: speed uses distance, velocity uses displacement. The cat example gives 0.4 m/s and 0.2 m/s, which are not equal.
  • Wrong: giving an average velocity with no direction. Right: average velocity is a vector, so write the direction too, for example 0.2 m/s to the left.
  • Wrong: dividing a winding-road length by time and calling it average velocity. Right: a winding road length is a distance, so distance ÷ time gives average speed.
  • Wrong: thinking zero average velocity means zero average speed. Right: after one full lap, displacement is 0 but distance is not, so average speed is not 0.

Practice

Question 1. Goumao (the cat) runs around at home: 2 m to the right, then 6 m to the left, 20 s in total. What are its average velocity and average speed?

Show the answer

Average velocity 0.2 m/s to the left; average speed 0.4 m/s. Taking right as positive, the displacement is 6 − 2 = 4 m to the left, so 4 m ÷ 20 s = 0.2 m/s. Average speed = (2 + 6) m ÷ 20 s = 8 m ÷ 20 s = 0.4 m/s. The two are not equal.

Question 2. After an earthquake, roads and bridges were wrecked and vehicles could not get through. To reach the disaster area quickly, a rescue team crossed obstacle after obstacle on foot, covering 60 km of rugged mountain road in 5 h. Which statement about the rescuers' velocity and speed is correct? ( )
A. The average velocity is 3.3 m/s
B. The velocity at every instant is 3.3 m/s
C. The average speed is 3.3 m/s
D. The speed at every instant is 3.3 m/s

Show the answer

C. The 60 km follows a winding mountain road, so it is distance, not displacement. Distance ÷ time is average speed: 60 000 m ÷ (5 × 3600 s) ≈ 3.3 m/s. The problem gives only total distance and total time, so D is wrong.

Words in English and Chinese

average velocity
平均速度
average speed
平均速率
instantaneous velocity
瞬时速度
instantaneous speed
瞬时速率
displacement
位移
distance (path length)
路程
uniform linear motion
匀速直线运动
tangent
切线

Sum it up before you move on.

Write this lesson in one line:

  1. What was the question?
  2. What is the simple version?
  3. What would you have got wrong?
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