Concepts
Quantities relating to motion are often vectors, but scalars can
be used in the case of straight-line motion, or if the direction is
not important in a particular situation. Sometimes different
names are used to distinguish them.
Scalar
Vector
How far?
distance
displacement m
How fast?
speed
velocity
How is the
speed changing?
acceleration acceleration ms-2
Unit
ms-1
Instantaneous and average values
The formula speed = distance / time is a simplification. It is
important to distinguish between an average and an instantaneous
value.
Average speed (velocity)
Average speed =
total distance
total time
Instantaneous speed (velocity)
Instantaneous speed = rate at which distance is changing with
time. This can be found from the formula speed = distance / time,
for a time which tends towards zero, or from the gradient of a
distance-time graph at a point in time. If the graph is a curve, than
the gradient of the tangent at the point in time gives the value.
The same principle applies to velocity (= displacement / time) and
acceleration ( = change in velocity / time)
Motion Graphs
velocity
ms-1
10
5
2
4
6
8
10
time / s
Sketch a-t and d-t graphs for this motion.
Interpreting Motion Graphs (gradient)
Section 1 of the Graph
The velocity increases from 0 to 10 ms-1 in a time of 2s. We can calculate the average
acceleration, but the fact that this section is a straight line shows that the acceleration is
constant.
a = (10 - 0) / 2 = 5 ms-2
This value applies at every point from 0s to 2s.
Section 2 of the Graph
In this section the velocity is not changing. a = 0 for the whole time.
Section 3 of the Graph
A constant negative gradient shows a constant negative acceleration.
a = (0 - 10) / 3 = -3.3 ms-2
Notice that the velocity is still positive.
Section 4 of the Graph
Again, the velocity is not changing. Its value of 0 is still a constant.
Again a = 0.
The gradient of the v-t graph gives the acceleration;
the gradient of the d-t graph gives the velocity.
Interpreting motion graphs (area)
Section 1 of the Graph
The average speed during the first 2s is 5 ms-1.
So the distance covered is 5 ms-1 x 2s = 10m. This corresponds to the area of the shape
under the graph line.
Section 2 of the Graph
Here the speed is a constant 10 ms-1 for a time of 5s. The distance covered is 50m. Note
that this is an additional 50m to the 10m covered in the first section. When deriving a d-t
graph from a v-t graph, make sure to add the values on to the previous total.
In general, the area of a trapezium
under the graph corresponds to the
formula for average speed. A triangle
or rectangle are special cases.
velocity
v
u
s = (u + v) t
2
Δt
time
(u+v)/2 = average speed and also the average of the parallel sides
Δt = distance between the parallel sides and also the time.
The area under a section of a v-t graph represents the additional
distance covered during the time. The area under a section of an
a-t graph represents the change in velocity during that time. Note
that areas below the time axis are negative and represent
negative displacements or velocity changes. In summary:
Uniform Acceleration
There are many situations in which an object is subject to uniform acceleration. i.e. an
acceleration which is constant in magnitude and direction. This is the case when the body
is subject to a constant force, such as gravity. A series of equations, based on the work
of Newton, relate the various kinematic quantities. These are commonly referred to by
students as the "suvat" equations, from the common symbols for the variables involved:
s = displacement (m)
u = initial velocity (ms-1)
v = final velocity (ms-1)
a = acceleration (ms-2)
t = time (s)
A v-t graph for a body
moving with uniform
acceleration looks like this:
The "suvat" Equations
From the gradient:
a = (v-u) / t
or v = u + at
From the area under the graph
(Area of a trapezium):
s = (u+v) t
2
Newton's equations
for uniformly
accelerated motion.
From the area under the graph
(Rectangle plus Triangle):
s = ut + ½(v-u) t
but v-u = at
so s = ut + ½at2
Squaring the first equation:
v2 = (u+at)2
= u2 + 2uat + a2t2
= u2 + 2a(ut + ½ at2)
so v2 = u2 + 2as
Free Fall and Air Resistance
Acceleration of Free Fall
The force of gravity on a body near the Earth's surface causes an acceleration of
approx. 9.8 ms-1 (providing air resistance is negligible). This is given the symbol g as it
also describes the gravitational field strength. (See Topic 6)
Air resistance
When a body moves through a fluid (liquid or gas) fluid friction acts in the direction
opposite to the velocity of the body. In the common case of a solid body moving
through the air, this is called "air resistance." The magnitude of this force depends on
the shape of the body and its velocity.
If a body is falling, the downward force of gravity is constant, but the upward force
of air resistance increases as the velocity increases. Eventually a velocity will be
reached where the air resistance equals the force of gravity. With a net force of
zero, there is no net acceleration.
The velocity at which this occurs is called the "terminal velocity."
Relative Velocity
If a body is moving with a constant velocity, this velocity appears different to a
stationary observer and one who is also moving. In all cases (except close to the
speed of light) the relative velocity of the observed body is equal to its absolute
velocity (as observed from a stationary point) minus the absolute velocity of the
observer.
vr = v - vo
Relative velocity in 1-D
Since the only directions are forwards and backwards, it is usual to use + for
forwards and - for backwards. Algebraic calculations will give relative velocity. e.g.
You are driving at 20 ms-1 along a straight road. A car approaches from the opposite
direction with a speed of 15 ms-1 (so -15 ms-1). The speed of the approaching vehicle,
relative to you is:
velocity of the other car - your velocity
= (-15) - 20 = -35 ms-1
i.e. 35 ms-1 in the direction you are coming from
Relative velocity in 2-D
In this case vector subtraction must be used. It is often useful to use the notation vAB
for the velocity of object A relative to B.
The general vector sum vAB = vAF + vFB, where F is a fixed point of reference can be
used. (vFB = -vBF)
Example
Car 1 is travelling north at 17 ms-1 when car 2 crosses an intersection ahead at 10 ms-1
from W to E without giving way. Find the velocity of car 2 relative to car 1.