Should velocity be defined as the rate of change of displacement?

What is displacement?

According to Fundamentals of Physics (10th edition) by D. Halliday, R. Resnick and J. Jewett:

The displacement of a particle is the change in its position.

Mathematically, displacement, x = final position - initial position and it is a vector quantity.

What is velocity?

According to Fundamentals of Physics (10th edition) by D. Halliday, R. Resnick and J. Jewett,

Velocity is the derivative of x (position) with respect to t (time).

In other words, velocity is the time rate of change of  position and it is also a vector quantity. Mathematically, velocity = dx/dt.

However, according to Physics Matters GEO 'O' Level Physics (4th edition) by C. Chew, S.F. Chow and B.T. Ho:

Velocity is the rate of change of displacement.

In other words, Physics Matters GEO 'O' Level Physics is saying that velocity is the rate of change of change of position. In my humble opinion, the definition might be technically incorrect.

Since I am on the topic of displacement and velocity, let me end off by posing a challenge to you: For motions in 1 dimension, are you able to determine the speed-time (distance-time) graph from its velocity-time graph (displacement-time) graph? How about velocity-time graph (displacement-time) from its speed-time (distance-time) graph?

O-level Science (Physics) N04/2/4bii

An athlete runs a 100 m race in 12.5 s. Explain why, for part of the race, the athlete must have been running faster than the speed calculated in (a).

GCE O-level 5052 Science (Physics) 2004 November Paper 2 Question 1(b)

Answer: The runner accelerates from rest.

Proof: Consider the area under an arbitrary speed-time graph that starts from rest and the area under another speed-time graph that is travelling at a constant speed, which is equal to the maximum speed of the first graph. Both graphs stop at time t. Thus, the area under the first speed-time graph  = average speed x t while the area under the second speed-time graph = maximum speed x t. Since first area < second area, hence average speed < maximum speed.

Oil drop challenge

A car moving with constant acceleration drips oil on the road at a rate of one drop every second. The diagram below shows the position of the oil drops along the road.

(a) Calculate the average speed of the car between position A and position C.

(b) Calculate the instantaneous speed of the car when it was at B.

Answers: (a) 3 m/s; (b) 4 m/s

What is the difference between deceleration and negative acceleration?

According to Five Easy Lessons: Strategies for Successful Physics Teaching by R. Knight,

Students interpret a positive acceleration as always meaning "speeding up" and a negative acceleration as "slowing down", rather than associating the sign with the direction of the acceleration vector. This is a difficult idea to change, and for many students it becomes a serious difficultly when they get to Newton's second law.

The sign of the acceleration indicates the direction of the acceleration vector. However, it does not indicate how the speed of the object will change. When the acceleration vector is parallel to the velocity vector, the speed of the object will increase. Conversely, when the acceleration vector is anti-parallel (in opposite directions) to the velocity vector, the speed of the object will decrease.

According to Physics for Scientists and Engineers (6th edition) by R. Serway and J. Jewett,

The word deceleration has the common popular connotation of slowing down. We will not use this word in this text, because it further confuses the definition we have given for negative acceleration.

In other words, deceleration is NOT equivalent to negative acceleration.