*Before you read this, I suggest that you read* posts 16.4, 16.12, 16.13 and 17.2.

I walked 2 km, in a straight line, stopped and then walked another 2 km, in a straight line. How far was I from my starting point? All you can tell me is that I was somewhere within a circle of radius 4 km whose centre was the place where I started. I could have turned around after 2 km and returned to my starting point. I could have continued in the same direction and finished 4 km from my starting point. To know my exact position, you would need to know the directions in which I was walking as well as the distances.

This combination of distance and direction is called *displacement*. Because displacement involves a direction, as well as a number, it’s a vector (see post 17.2). Distance is the modulus (see post 17.2) of displacement.

*Velocity* is the rate of change of displacement. If Δ* r* is the change in displacement that occurs in a time-period Δ

*t*, the

*average velocity*in this time period is

**v**_{av} = Δ* r*/Δ

*t*.

Here we are following the convention of using the symbol Δ to mean a part of something, as in post 16.12. *Speed* is the modulus (see post 17.2) of velocity, so the *average speed* is defined as

*v** _{av}* = Δ

*r*/Δ

*t*

where Δ*r* is distance. This relationship between velocity and speed was explained, less mathematically, in post 16.2. Both velocity and speed are measured in km.s^{-1} (km/s) (see post 16.12).

If the time interval is very short, the part of the curve used to calculate speed will be very close to a straight line. We use the symbol δ to mean a tiny part (almost zero) of something. The picture shows that, in the time-period δ*t*, the red curve is almost a straight line; in this time-period the average speed is almost constant and so is approximately equal to the true speed. We can express this idea by defining the approximate true speed by

*v* ≈ δ*r*/δ*t*

where the symbol ≈ means “approximately equal to”.

What is the exact definition of *true speed*? The closer we make δ*t* equal to zero, the better our approximate definition becomes. When δ*t* is just about to vanish, the definition becomes exact. We call this point the *limit* of δ*r*/δ*t* as δ*t* tends to zero. Then the definition of true speed can be written as

*v* = Limit(δ*r*/δ*t*)

δ*t → *0

This way of writing the limiting value of a little part of one thing divided by a little part of another is very clumsy. So, we usually write the definition of true speed by

*v* = *dr*/*dt*.

Be careful! The symbol *d*/*dt* represents a mathematical operation (*differentiation with respect to time*) performed on *r*. So you must **never** treat this *d* as something that represents a number – so, for example you can’t cancel the top and bottom of the right-hand side of the equation above by *d* to get *v* = *r*/*t*.

By analogy, the definition of true velocity is * v* =

*d*

*/d*

**r***t*.

*Acceleration* is the rate of change of velocity. So, we can define acceleration by

* a* = d

**/d**

*v**t*.

Because * v* is a vector, something that changes direction without changing its speed is still undergoing a change in velocity, so that it is accelerating. So, an object that is moving in a circular path at a constant speed is constantly accelerating. Acceleration is measured in the units of velocity divided by the units of time – (km/s)/s = km/s

^{2}or km.s

^{-2}(see post 16.13).

To calculate acceleration from displacement, we perform the operation of differentiation with respect to time twice. The first time we convert a displacement into a velocity. The second time we convert the velocity into an acceleration. We represent this double differentiation mathematically by

* a* = d

**/d**

*v**t*= d

^{2}

*/d*

**r***t*

^{2}.

The important point to make is that displacement, velocity and acceleration are all vector quantities. If all this stuff about differentiation looks a bit abstract, I have tried to make it more concrete in the appendix below.

*Related posts*

17.3 Three-dimensional vectors

17.2 Vectors

16.13 Changes in movement

16.12 Measuring movement

*Appendix – differentiation*

__Calculating velocity and acceleration from displacement__

I’m going to think about an object moving, on a flat surface, whose displacement (measured in metres) is described by

* r* = 5.0

*t*

**i**^{2}+ 6.0

*j**t*

where *t* represents the time the object has been moving (measured in seconds). The vector ** i** is a unit vector (a vector whose length is the number 1) pointing in the direction of an

*X*-axis (see post 17.2);

*is a unit vector in the*

**j***Y*-axis direction. Remember that

*t*

^{2}=

*t*×

*t*.

If *u* = *nt* and *w* = *mt*^{2}, where *n* and *m* are constants, I can tell you that *du*/*dt* = *n* and *dw*/*dt* =2m*t*. These statements are proved in the next sections of the appendix.

From what I’ve told you in the previous paragraph, we can calculate the velocity of the object as

* v* =

*d*

*r**/dt*= 10

*i**t*+ 6

*m.s*

**j**^{-1}.

So its speed is equal to the modulus of * v* which is equal to √(10

*t*×10

*t*+ 6

^{2}) (see post 17.2). When

*t*= 1 s, this speed is equal to √(10

^{2}+ 6

^{2}) = √(100 + 36) = √136 = 12 m.s

^{-1}(because 11.7 ×11.7 = 136 and 11.7 when measured with a precision of two digits is 12, see post 16.7). Its direction is the angle whose tangent is (6/10) (see post 17.2) and is equal to 31

^{o}to the

*X*-axis (see post 17.2).

Note that the component of * v* in the

*Y*-axis direction is a constant, so its acceleration in this direction is zero.

The acceleration of the object is

** a** =

*d*

**v***/dt*= 10

*m.s*

**i**^{-2}.

So the acceleration is a constant 10 m.s^{-2} in the direction of the *X*-axis.

__Proof that__*du*__/__*dt*__=__*n*

We defined *u* by the equation

*u* = *nt.*

When we increase the value of *t* by an increment δ*t* the new value of *u* is given by

*u* + δ*u* = *n*(*t* + δ*t*).

This result follows by simply replacing *u* by *u* + δ*u* and *t* by (*t* + δ*t*). Subtracting the first equation from the second gives

δ*u* = *n*(δ*t*).

Dividing both sides of the resulting equation by δ*t* gives

δ*u*/ δ*t* = *n*.

The right-hand side of this equation is the constant *n*. So, its value doesn’t change when δ*t* vanishes with the result that

*du/dt* = *n*.

__Proof that__*dw*__/__*dt*__=2m__*t*

We defined *w* by the equation

*w* = *mt*^{2}.

When we increase the value of *t* by an increment δ*t* the new value of *w* is given by

*w* + δ*w* = *m*(*t* + δ*t*)^{2} = *m*(*t*^{2} + 2*t*δ*t* +(δ*t*)^{2}).

The first part of this result follows by simply replacing *w* by *w* + δ*w* and *t* by (*t* + δ*t*). If you don’t understand the second part, see the appendix below. Subtracting the first equation from the second gives

δ*w* = *m*(2*t*δ*t* +(δ*t*)^{2}).

Dividing both sides of this equation by δ*t* gives

δ*w/**δ**t* = *m*(2*t* +δ*t*) = 2*mt* + *m*δ*t*.

In the limit as δ*t* → 0, this becomes

*dw/dt* = 2*mt*.

__Proof that (__*a*__+__*b*__)__^{2}__=__*a*^{2}__+ 2____ab____+__*b*^{2}__.__

(*a* + *b*)^{2} = (*a* + *b*) × (*a* + *b*) = [*a* × (*a* + *b*)] + [*b* × (*a* + *b*)] = *a*^{2} + *ab* +*b*^{2} + *ba*.

Since *ab* = *a* × *b* and *ba* = *b* × *a*, *ab* = *ba*. So the right-hand side of the line above becomes

*a*^{2} + *ab* +*b*^{2} + *ba*. = *a*^{2} +*ab* +*b*^{2} + *ab* = *a*^{2} + 2*ab* +*b*^{2}.

To satisfy yourself that this is true consider what happens when *a* = 2 and *b* = 3. Then

(*a* + *b*)^{2} = 5 × 5 = 25 and *a*^{2} + 2*ab* +*b*^{2}. = 4 + (2 ×2 × 3) + 9 = 25.