For a particle moving in 3-dimensional space, the velocity is the
time-derivative of the position vector r.
Derivatives of vectors are defined as limits, in the same way as
in ordinary calculus. The standard rules of calculus apply.
Since r = xi + yj + zk, differentiating gives
dr
dx
di dy
dj dz
dk
=
i+x + j+y + k+z
dt
dt
dt
dt
dt
dt
dt
We use fixed axes, so i, j, k are constant, and have zero
derivatives. So half the terms in the equation above vanish,
giving
ṙ(t) =
dr
dx
dy
dz
=
i+
j+
k = ẋ i + ẏ j + ż k
dt
dt
dt
dt
(1.25)
Here we use Newton’s notation: a dot over a letter means
differentiation with respect to time.
Equation (1.25) means: to differentiate a vector, you just
differentiate each component, provided that
Example: Uniform circular motion. Take positive constants a
and ω (Greek omega, ω, not the ordinary Roman letter w). Then
r(t) = a cos ωt i + a sin ωt j
(1.26)
describes a point P going round a circle of radius a in the xy
plane with centre at the origin O. The vector r = OP is at angle
ωt to the x axis. As t increases, P rotates anticlockwise, and it
takes a time 2π/ω to go round once (since that is the time it
takes for ωt to increase by 2π).
The time taken to go round once is called the period, and ω is
called the angular speed of rotation: the number of radians per
second.
In general the velocity of a moving particle is always along the
tangent to the path. This is clear from a diagram.
Note: speed versus velocity
Speed: a scalar, distance travelled per unit time.
Velocity: a vector, in the direction of travel, with
magnitude equal to the speed.
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Parametric and Cartesian Descriptions
Equation (1.26) defines a curve in the xy plane: a circle. In
general, an equation of the form (1.24) describes a curve in
space; it is called a parametric description of the curve, with
parameter t.
The parametric equations (1.26) can be converted to a
conventional (Cartesian) equation for the curve by writing
(1.26) in components, and eliminating the parameter t. We have
x = a cos ωt,
y = a sin ωt,
z=0
It is easy to eliminate t by squaring and adding the x and y
equations, giving
x 2 + y2 = a2 ,
z=0
This is the Cartesian equation of a circle of radius a in the xy
plane.
There are other parametric forms for the circle: the parameter
does not have to be time. For example,
x = a sin φ,
y = a cos φ,
z=0
is another parametric form for the same circle. This equation
gives the coordinates of a general point on the circle, with
parameter φ. The point corresponding to φ = 0 is r = aj. As φ
increases, the point moves round the curve clockwise.
1.4.3
Acceleration
The acceleration vector is defined as the time-derivative of
velocity. So it is the second derivative of r. Second derivatives
are denoted by two dots. So we have
r̈(t) = ẍ i + ÿ j + z̈ k
(1.27)
Simple Example: Motion along a Line. If the particle is
moving along the x-axis, we have
r(t) = x (t)i
It follows that ṙ = ẋ i and r̈ = ẍ i.
The acceleration and velocity are parallel to r. This is obvious
intuitively as well as mathematically.
But in general the acceleration and velocity are not parallel.
Example: Uniform Circular Motion. In the last section we
considered a particle going round a circle of radius a with
angular speed ω:
r(t) = a(cos ωt i + sin ωt j)
21
The acceleration is
r̈(t) = − aω 2 (cos ωt i + sin ωt j) = −ω 2 r(t)
(1.28)
The acceleration is not in the same direction as the velocity.
It is in the same direction as the position vector. But that is not a
general rule; it happens here only because this example is so
simple.
Example: General Circular Motion A particle P goes round
the same circle as in the example above, but not at a constant
rate.
We describe the motion by the angle θ between OP and the
x-axis, where θ is a given function of t. In the example above
θ (t) = ωt. Now θ can be any function of t.
We have
r(t) = a cos θ (t) i + a sin θ (t) j
(1.29)
The chain rule of calculus gives
d(cos θ )
d(cos θ ) dθ
=
= − sin θ θ̇,
dt
dθ
dt
where as usual the dot stands for the time-derivative, so
θ̇ = dθ/dt.
Using this and a similar expression for the derivative of sin θ,
we have
ṙ(t) = − a sin θ (t) θ̇ i + a cos θ (t) θ̇ j = aθ̇ (− sin θ i + cos θ j)
Note
The acceleration is
r̈ = aθ̈ (− sin θ i + cos θ j) + a θ̇ (− cos θ i − sin θ j) θ̇
This can be written
where
θ
r̈ = − aθ̇ 2 br + aθ̈ b
br = cos θ i + sin θ j and b
θ = − sin θ i + cos θ j.
(1.30)
(1.31)
It is easy to see that br and b
θ are unit vectors. The vector br points
in the radial direction, directly away from O. The vector b
θ is
22
perpendicular to br. The direction of b
θ is tangential to the circle,
in the direction corresponding to increasing θ.
The acceleration (1.30) has a component along the circle if the
speed of the particle is not constant. This often called the
tangential component. The first term in (1.30) is the radial
component, pointing towards the centre O of the circle.
1.4.4
Integration: Deducing Position from Velocity
So far we have been deducing the velocity and acceleration by
differentiating the position vector. Now suppose we are given
the velocity as a function of time. Then the position is found by
integration.
Position and Velocity in One Dimension
dx
In elementary calculus, if v =
then x =
dt
constant of integration.
Z
v dt and there is a
Example: Suppose a particle moves along the x-axis with
velocity given by
ẋ (t) = αt2
at time t, where α is a given constant.
Then
Z
αt3
+C
3
where C is an unknown constant. Can we find C?
x (t) =
αt2 dt =
It’s obvious that given something’s velocity, you can’t deduce
its position unless you’re given some extra information (in
plain language: if you know how fast something is going, it
doesn’t tell you where it is).
If you’re given the position at some initial instant, say t = 0,
then you can find its position for all future times.
Suppose you are given that the particle is at the point X of the
x-axis at t = 0. Then
Position and Velocity in Three Dimensions
For vectors in general, if A =
dB
then B =
dt
23
Z
A dt.
This integral is evaluated using components, as for
differentiation. In fact, if
A(t) = A1 (t) i + A2 (t) j + A3 (t) k and A =
dB
dt
(1.32)
then
B=
Z
A dt =
Z
A1 (t) dt i +
Z
A2 (t) dt j +
Z
A3 (t) dt k (1.33)
There is a constant of integration in each of the three integrals.
Example. Suppose a particle has velocity
v(t) = ai + b cos t j + b sin t k
at time t, where a and b are constants. Then its position at time t
is
r(t) = ( at + C1 )i + (b sin t + C2 )j − (b cos t + C3 )k
where C1 , C2 , C3 are unknown constants of integration.
To find them, we must be given the position of the particle at
some time.
Suppose the particle is at the origin at time t = 1. Putting t = 1
into the equation above gives
The orbit is a helix.
The difference between a helix and a spiral is
1.4.5
Deducing Position from Acceleration
The relation between (1.32) and (1.33) is a general one. If we
take B to be velocity, then A is acceleration. If the acceleration is
given, you can find the velocity by integration, and then the
position by integrating again.
When integrating the acceleration to deduce the velocity, an
initial velocity must be given to determine the constants of
integration. Then the velocity is completely determined, and
24
the position can be found by integrating again, provided an
initial position is given..
Example: constant acceleration. This is a very simple case:
r̈ = a, where a is constant.
Since no axes have been specified, we can choose them in any
direction we like. It makes life easier to take one of the axes in
the direction of the acceleration vector. We take the z-axis
parallel to a, then
r̈ = ak
where a = |a|.
Integrating, we have
To determine the constants, we must be given an initial velocity.
Suppose ṙ(0) = V0 where V0 is given. Then
ṙ = V0 + atk
Now integrate again, remembering that V 0 and a are constant:
r(t) = V0 t + 12 at2 k + C0
where C0 is another constant of integration. To find it, we must
be given the initial position of the particle.
Suppose r(0) = R0 where R0 is given. Then
r(t) = R0 + V0 t + 12 at2 k
(1.34)
A nice formula for the position vector of a particle with
acceleration a in the z direction, when the particle starts at the
point R0 with velocity V0 at time 0.
Specific Example. Suppose, for simplicity, that the particle
starts at the origin (so R0 = 0), with initial velocity
V0 = V cos θ i + V sin θ k
This means that the initial velocity has magnitude V and is at
angle θ to the acceleration vector.
Substituting into (1.34) gives
25
y = 0,
z = x tan θ +
ax2
2V 2 cos2 θ
(1.35)
z is a quadratic in x, so the orbit is a parabola in the plane y = 0.
We will see later that the orbit of a ball through the air is a
parabola, because it moves with constant acceleration.
Question: We assumed that the initial velocity V had no y
component. To get a general form for a a constant-acceleration
orbit, should we rework the calculation with a general V?
1.5 Relative Velocity
1.5.1
Basic Ideas
Relative Position
We defined the position vector of a particle P with respect to an
origin O to be OP. We can choose O in any way we like.
Now suppose we have two particles, P1 and P2 , with position
vectors r1 , r2 with respect to an origin O. We define the position
vector of P2 relative to P1 by
r2 rel 1 = r2 − r1 .
(1.36)
This is just the position vector of P2 with respect to an origin at
the position of P1 . It tells you how far P2 is from P1 , and in what
direction.
Obviously
r1 rel 2 = −r2 rel 1
Relative Velocity
This is defined in the obvious way:
v2 rel 1 =
dr2 rel 1
= ṙ2 − ṙ1
dt
So
v2 rel 1 = v2 − v1
26
(1.37)
Here v1 and v2 are the velocities of P1 and P2 with respect to an
origin O. It doesn’t matter what origin is used.
The meaning of relative velocity is as follows. If you were
attached to P1 and moving with it, then P2 would appear to be
moving with velocity v2 rel 1 .
Simple Example. A small plane is flying due west at a speed
of 60 miles/hour (mph) relative to the ground. Suppose it is
flying directly into a headwind of 30 mph. How fast is the
plane going relative to the wind?
“Headwind” means the wind is blowing in the direction
opposite to the plane’s velocity; eastwards in this case.
This is a trivially easy example, and the answer should be
obvious. We solve it using vectors, to give practice in the
technique in preparation for harder examples.
Let i be a unit vector pointing East.
The velocity of the wind relative to the ground is v w = 30i.
The velocity of the plane relative to the ground is v p = −60i.
We want to find the velocity of the plane relative to the wind.
This is given, according to (1.37) by
v p rel w = v p − vw = −60i − 30i = −90i
(1.38)
So the plane is flying through the air at a speed of 90 mph.
What use is this answer? Well, the maximum speed of a plane is
the maximum speed that it can fly relative to the air (think
about it). So if this (very slow) plane has a maximum speed of
90mph, that means that in a headwind of 30 mph, its maximum
speed over the ground is 60mph.
1.5.2
Examples
Example 1. A small plane is flying relative to the ground at a
speed of 90 mph in a northwesterly direction, i.e., 450 west of
north. A steady wind of 30 mph is blowing from the west. How
fast is the plane flying through the air?
Take unit vectors i eastwards and j northwards.
Then the wind velocity is
vw = 30i.
27
The velocity of the plane has magnitude 90 and is at an angle
3π/4 to i, and π/4 to j. So
90
90
v p = 90 cos(3π/4)i + 90 cos(π/4)j = − √
i+ √
j
2
2
So the velocity of the plane relative to the wind is:
v p rel w = v p − vw =
90
90
− √ − 30 i + √ j
2
2
(1.39)
The speed of the plane through the air is the magnitude of this:
|v p rel w | =
s
90
− √ − 30
2
2
+
8100
≈ 113.2mph
2
The direction of v p rel w is at angle θ West of North, defined by
v p rel w · j = |v p rel w | cos θ
(1.40)
√
and v p rel w · j is the j component of v p rel w , which is 90/ 2
from (1.39).
So
cos θ =
90
90
√ ≈√
≈ 0.56
|v p rel w | 2
2113.2
Hence θ = ±55.8◦ or ±0.98 radians. Which sign should we
take?
Equation (1.39) shows that the relative velocity has a negative i
component. Since i points East, it follows that v p rel w is West of
North. So θ = 55.8◦ W of N.
What is the meaning of this angle?
Approach by a diagram
28
Example 2. A plane has a maximum speed of 200mph through
the air. It is to fly 100 miles South-West relative to the ground,
in a wind of 50mph blowing Northwards. What is the
minimum time for the journey?
Here neither v p nor v p rel w are known; you know the direction
of v p and the magnitude of v p rel w which is 200 since the plane
must go flat out in order to achieve the minimum time.
Solution:
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