17.4 Displacement, velocity and acceleration

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 Δr is the change in displacement that occurs in a time-period Δt, the average velocity in this time period is

vav = Δrt.

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

vav = Δrt

where Δr 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 δt, 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 ≈ δrt

where the symbol ≈ means “approximately equal to”.

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

v = Limit(δrt)
δt → 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 = dr/dt.

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

a = dv/dt.

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/s2 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 = dv/dt = d2r/dt2.

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

Follow-up posts

22.10 Differentiation

Appendix – differentiation

  1. 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 = 5.0it2 + 6.0jt

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); j is a unit vector in the Y-axis direction. Remember that t2 = t × t.

If u = nt and w = mt2, where n and m are constants, I can tell you that du/dt = n and dw/dt =2mt. 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 = dr/dt = 10it + 6j m.s-1.

So its speed is equal to the modulus of v which is equal to √(10t ×10t + 62) (see post 17.2). When t = 1 s, this speed is equal to √(102 + 62) = √(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 31o 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 = dv/dt = 10i m.s-2.

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

  1. Proof that du/dt = n

We defined u by the equation

u = nt.

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

u + δu = n(t + δt).

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

δu = nt).

Dividing both sides of the resulting equation by δt gives

δu/ δt = n.

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

du/dt = n.

  1. Proof that dw/dt =2mt

We defined w by the equation

w = mt2.

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

w + δw = m(t + δt)2 = m(t2 + 2tδt +(δt)2).

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

δw = m(2tδt +(δt)2).

Dividing both sides of this equation by δt gives

δw/δt = m(2tt) = 2mt + mδt.

In the limit as δt → 0, this becomes

dw/dt = 2mt.

  1. Proof that (a + b)2 = a2 + 2ab + b2.

(a + b)2 = (a + b) × (a + b) = [a × (a + b)] + [b × (a + b)] = a2 + ab +b2 + ba.

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

a2 + ab +b2 + ba. = a2 +ab +b2 + ab = a2 + 2ab +b2.

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

(a + b)2 = 5 × 5 = 25 and a2 + 2ab +b2. = 4 + (2 ×2 × 3) + 9 = 25.


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