Kinematics is the study of motion without reference to mass or force. It is concerned with problems such as predicting the position of an object at some future time, given its initial position and velocity.

5m

Recall that a scalar quantity is one that just has a size (a number plus a unit, e.g. 10m) whereas a vector quantity has a direction as well as size (e.g. 10m to the left). So,

- distance is a scalar quantity
- displacement is a vector quantity

A unit of speed comes from whatever units of distance and time are used
in the equation.

Example: If a car moves 100km in 2 hours, its average speed is 100km/2h = 50km/h.

The SI unit for speed is m/s, which we also write as
ms^{-1}.

**Example**

An athlete does the 100m sprint in 10s. Find his average speed. Why do we say ‘average’ speed?

Average speed = distance/time = 100m/10s = 10m/s.

We say ‘average’ speed because the athlete starts from rest and accelerates, so his speed is changing as he runs – but he has an average speed over the 100m.

From the definition, we see that speed and velocity have the same units.

Note: We often use the terms 'speed' and 'velocity' as if they were the same, but in physics they are strictly speaking different, since.

- speed is a scalar quantity – it has size only (i.e. a number
and a unit E.g. 10ms
^{-1}) - velocity is a vector quantity – it has size
*and direction (*E.g. 10ms^{-1}to the right)

A car travels along the path AXYZB, a distance of 1000m, in 100s. The straight-line distance AB is 500m.

- average speed = distance/time = 1000m/100s = 10m/s
- average velocity = displacement/time = 500m/100s = 5m/s in the direction from A to B

An athlete runs exactly once round a 400m track in 60s. What is his (a) average speed, (b) average velocity?

a) average speed = distance/time = 400m/60s = 6.7 m/s

b) average velocity, symbol <v> = displacement/time =
0m/60s = 0

Displacement is the straight line distance from start to finish, and since he starts and finishes at the same point his displacement is zero, therefore his average velocity is zero. However, his instantaneous velocity at any given point is not zero:

**Acceleration, a**

Any velocity unit divided by any time unit is a possible unit
for acceleration, so we could have the unit ‘miles per hour per second’ or ‘km/h
per second’ etc. However, the standard unit, i.e. the* *SI unit, is ‘m/s
per second’. This can be expressed as (m/s)/s or ms^{-1}/s or m/s^{2} or ms^{-2} (the last two being said as ‘metres per
second squared’).

A familiar acceleration is that produced by gravity. We know
that if we drop something it usually falls towards the ground, getting faster as
it falls. It is therefore accelerating towards the ground. Since it has a
direction, *acceleration is a vector quantity*.

By experiment, we find that if an object is released from rest, and if air resistance is neglected, then:

- after 1 second it reaches 10 ms
^{-1}(to be more exact, 9.8 ms^{-1}) - after 2 seconds it reaches 20 ms
^{-1} - after 3 seconds it reached 30 ms
^{-1}etc.

We say that it accelerates downwards at a rate of 10 ms

This is called the ‘

Thus: g = 10 ms

The definition of acceleration means that something is
accelerating if its velocity changes, and since velocity includes speed
*and* direction, then *something is accelerating if its speed, its
direction or both change.*

Consider a cork spun at a constant speed on a thread:

The direction of the velocity is always at a tangent to the curve, and so is different at every point, and so the velocity is constantly changing in direction, though not in magnitude, so the cork

If you perform the above experiment, you will feel a constant pull on your hand. This is because the force on the cork is provided by the tension in the thread. Thus, the force on the cork is directed from the cork towards the centre of the circle. Newton’s laws are discussed later, but recall from GCSE that F = ma, so acceleration has the same direction as force, so the cork in the above example is constantly accelerating towards the centre of the circle.

**LINEAR MOTION **(return
to start of page)

**Sign convention**

There are only two directions along a given straight line, so we can indicate the directions of any vectors along a straight line simply by choosing one direction as positive and the other as negative.

**Example**

An object is projected straight up at 30ms^{-1}. Using g = 10ms^{-2}, and neglecting air resistance, what will
be its velocity:

i) after 2 sec, ii) after 4
sec?

i) Let’s choose the sign convention that ‘up is positive’. Thus:

The initial velocity is positive because its direction is upwards. The
acceleration is negative because the acceleration due to gravity always acts
downwards.

We can do this particular problem without
specifically using equations:

When an object falls with an acceleration of g
=10ms^{-2}, it speeds up by
10ms^{-1 }each second. When an
object is projected upwards, it slows down by 10ms^{-1
}each second.

Thus, after 2s the speed is 10ms^{-}^{1}

The negative sign means that the velocity is directed downwards, so the
object must have reached the top of its motion, and is now falling to
earth.

**Distance-time graphs (or s-t graphs)**

**a) Zero velocity**

A horizontal line means that distance is unchanging, i.e. the object is at rest, i.e. the speed or velocity is zero.

**b) Constant velocity**

A straight, inclined line means that the object is moving with constant velocity.

The slope (or gradient) of the line equals the velocity. To find the slope:

- draw a triangle such as shown above
- work out A and B
*from the axes*(don't measure them in cm) - slope = A/B

**c) Changing velocity**

A curve, as above, means that the velocity is changing.

We can find the velocity at any instant time, such as at time t, called the ‘instantaneous velocity’ at that time:

- find the point on the graph corresponding to time t, labelled P above
- draw the tangent to this point – this is a line that touches the curve at just this point, but does not cross the curve
- complete a triangle as shown, and calculate the slope (=A/B) – this equals the velocity at the instant of time, t

The following is a distance-time graph for an object. Describe its motion.

There are no numbers on the graph, so we cannot determine actual values.
However, the slope of the graph at any point equals the velocity at the
corresponding time.

- At time t
_{2}the slope, and therefore the velocity, is zero - At times such as t
_{1}before t_{2}the slope, and therefore the velocity, is always positive, though not constant - At times such as t
_{3}after t_{2}the slope, and therefore the velocity, is always negative, though not constant

**Velocity-time graphs (or v-t graphs)**

**a) Constant velocity**

A horizontal line means that the velocity is constant, which means that the acceleration is zero.

**b) Changing velocity at a uniform (i.e. constant)
rate**

In both the above cases the acceleration equals the slope of
the graph.

In the right hand case, the object is
slowing down, i.e. decelerating, as indicated by the slope, and therefore the
acceleration, being negative.

**c) Area below v-t graphs**

*The area below any section of a v-t graph equals the distance travelled in that section*

The above is a velocity-time graph for a certain object. It is made up of three sections: A, B, C.

- determine the acceleration in each section
- determine the distance travelled in each section and the total distance travelled
- describe the motion of the object

- Section A: acceleration, a = (v -u)/t = (10 – 0)/2 = 10/2 =
5 (m/s)/s = 5 m/s
^{2} - Section B: acceleration, a = zero as the graph is horizontal
- Section C: acceleration, a = (v – u)/t = (0 – 10)/ 4 = -10/4
= -2.5 m/s
^{2}

- Section A: distance = area of triangle = ½ base x height = ½ x 2 x 10 = 10 m
- Section B: distance = area of rectangle = 4 x 10 = 40 m
- Section C: distance = ½ base x height = ½ x 4 x 10 = 20m
- Total distance = 10 + 40 + 20 = 70 m

- The object starts from rest
- It accelerates for 2 sec at 5m/s
^{2 }reaching 10 m/s, moving 10m - It continues at 10 m/s for 4 sec, moving a further 40m
- Then it decelerates at a rate of 2.5 m/s
^{2}coming to rest in 4 sec, moving a further 20m. - It moves a total distance of 70m

**a) Distance travelled, s**

Suppose that an object moving along a straight line accelerates uniformly (i.e. at a constant rate) from an initial velocity u to a final velocity v in time t.

Use the above equation to work out the distance moved in each of the 3 section in the last example:

- Section A: distance, s = ½ (0 + 10) x 2 = ½ x 10 x 2 = 10m
- Section B: distance s, = ½ (10 + 10) x 4 = ½ x 20x 4 = 40m
- Section C: distance s, = ½ (10 + 0) x 4 = ½ x 10 x 4 = 20m

**c) Equations of motion**

There are three equations of motion which apply to an object moving with uniform acceleration. The first we have already seen:

**Derivations**

**1.** The first equation, v = u + a t, comes directly from
the definition of acceleration, a = (v – u)/t

**2. **To get the second equation, we can use:

**3.** For the third equation, we have:

**Example**

A car starting from rest accelerates at 2 m/s^{2 }for 5 sec.

- What velocity does it reach?
- How far does it travel in the 5 sec?

a) v = u + a t = 0 + 2 x 5 = 10 m/s

b)
s = u t + ½ a t^{2} = 0 + ½ x 2 x 5^{2} = 25 m

**Example**

1) The trolley moves a distance of 193cm from rest before hitting the wooden block at the end of the runway. The time taken is 3.4 seconds. Calculate: i) the acceleration ii) the velocity reached

2) The trolley hits the block and the spring-loaded plunger makes it rebound. It rebounds 102cm before stopping. If its deceleration had the same magnitude (size) as its acceleration, what speed did it bounce off at?

1) i) s = 193cm, u = 0, t = 3.4 sec

We want to find acceleration a.

2) Deceleration is the same a negative acceleration.

So when the trolley moves down the slope, its acceleration, a
= +33.4cm/s^{2}.

When it rebounds, its deceleration, or negative acceleration,
a = -33.4cm/s^{2}

**Example**

A body is moving at 4m/s and
accelerates at 1.5m/s^{2} for 6s.

- What velocity does it reach?
- What is the average velocity over the 6s?
- Sketch a v-t graph for the motion
- Use the graph to find the distance moved
- Use <v> = s/t to find the distance moved
- Use s = u t + ½ a t
^{2 }to find the distance moved

b)

c)

d)

e)

f)

**MOTION UNDER GRAVITY
**(return
to start of page)

An experiment believed to have been performed by Galileo Galilei (1564-1642):

Using balls of the same size means that the air resistance is the same on both of them. Thus, the experiment showed that objects of different masses accelerate at the same rate, if air resistance is negligible.Further experiments indicate that:

- At any place near the earth’s surface, all bodies falling
freely under gravity, in a vacuum or when air resistance is negligible,
accelerate at the same rate
*irrespective of their masses*

The size of the *'acceleration due to gravity' *or
*'acceleration of freefall' *on the earth is denoted by g and equals
9.8ms^{-2}. In problems we
usually use g = 10 ms^{-2}.

**Rough method of finding g**

The problem with the above method is the difficulty of measuring the time accurately. It would be better if the timing were automatic – as below:

The coil is wound around a soft-iron core, and together they form an electromagnet. So-called, 'soft-iron' is used for the core because it easily magnetises and demagnetises. When an electric current passes through the coil, the iron becomes a magnet and when the current is turned off, it demagnetises.In the position shown, the two-way switch completes the electromagnet circuit, and the steel ball is held as shown.

When the switch is flipped to the bottom position, the current in the electromagnet stops, so the ball starts to fall. At the same moment the bottom circuit is competed, so the clock starts.

The trap door is hinged, and when the ball hits it, it opens and cuts off the current to the clock, stopping the clock. Thus, the above lets us automatically time the ball falling though the distance s.

**Typical results**

distance, s = 1.69m

time to fall, t =
0.583s, 0.588s, 0.584s. So average time, t = 0.585

Note – taking several measurements, and finding the average, improves accuracy

We can use s = ut + ½ a t^{2}, but we replace ‘a’ by ‘g’, to find the
acceleration due to gravity, g.

**Calculations for motion under gravity**

Since the equations of motion apply to objects moving with a uniform (constant) acceleration, then they apply to bodies moving freely under gravity, since the acceleration due to gravity is constant (near the earth's surface).

**Example**

A stone is dropped and takes 2s to reach the ground. What is
(a) the velocity at impact (b) the height from which it was dropped (neglect air
resistance and take g = 10m/s^{2})

Letting downwards be positive, then acceleration, a = +g =
+10m/s^{2}

**Example**

A ball is projected vertically upwards with an initial velocity of 15m/s.

- What is its velocity at its maximum height? (no calculation needed)
- How long does it take to reach its maximum height?
- What is the maximum height?
- What is the time to fall back to the thrower?
- What speed does it arrive back at the thrower?

(a) The ball starts at 15m/s and slows as it rises. At the maximum height it has zero velocity for an instant, and then it start to fall downwards.

(d) We might guess that the ball takes the same time to fall as it took to rise. However, we can check this by calculation.

At the maximum height its speed is zero, so this is its initial
velocity for the downwards trip, and it has to fall 11.25m. We are free to
choose up or down as positive, so long as we are consistent within a given
calculation. Suppose that now we take down as being positive, then the
acceleration a = +g = +10m/s^{2}.

**Projectiles**

The diagram represents a multiflash image of two objects. One is dropped, and the other is projected sideways at the same moment.

Notice that

This indicates that:

*The vertical and horizontal motion of an object are independent, and can be treated separately*

A shell is fired sideways from an 80m high cliff at 30m/s.

- Sketch the approximate path followed by the shell
- How long will it take for the shell to impact the ground?
- What is the horizontal range of the shell – i.e. how far from the base of the cliff will the shell hit the ground?
- What is vertical component of velocity on impact?
- What is the velocity of impact (i.e. its speed and direction of motion)?

(a) The path will be a curve, as represented by:

(b) Though the shell is moving along a curve, we can think of it this being made up of a combination of vertical and horizontal motion, and treat each independently.

The shell starts with an horizontal speed of
30ms^{-1}, but it has zero
initial vertical speed.

So, taking down as positive, vertically: u = 0, s = +80m
and g = +10m/s^{2}

(c) The force of gravity only acts vertically – it has no effect on the
horizontal speed. Also, we are told to neglect air resistance. This means that
the shell moves at a constant speed of 30m/s horizontally for the 4sec that it
takes to reach the ground.

(d) For vertical motion, we can treat the shell as if it were simply
dropped (i.e. from rest) from a height of 80m:

(e) On impact, the shell has:

- a horizontal speed of 30ms
^{-1 }(= initial horizontal speed) - a vertical speed of 40 ms
^{-1 }(from (d))

Note: We often do not bother giving vectors special symbols,
especially if the vectors act on the same straight line. However, there are
various ways to represent a vector, such as **v** or __v__ . The magnitude
or size or length of **v** or __v__ would then be represented by
|**v**| or |__v__| or simply v.

**Example**

A projectile is fired from the ground with a velocity of
500ms^{-1} at
30^{0} to the
horizontal.

Calculate:

- the maximum height
- the time of flight
- the horizontal range

We can treat vertical and horizontal motion independently, so
firstly we determine the vertical and horizontal components of velocity:

Check
using Pythagoras: (433^{2} +
250^{2} )^{1/2}
= 500

a) Vertically the object rises with an initial speed of
250ms^{-1}. It slows as it
rises, and at its maximum height is momentarily at rest (in the vertical
direction).

So, v = 0; u = 250ms^{-1}; a = -10ms^{-2} ; s = s_{max}
= ?

b) t = time to rise to maximum height

The time to fall back to the ground would be the same, so the time of
flight is 2 * 25 = 50 sec

c) In the 50s that it is in the air, the object moves at 433
ms^{-1} horizontally.

(this is the same as using s = u t + ½ a t^{2} , with a = 0, since there is no
*horizontal* acceleration)

**RELATIVE MOTION **(return
to start of page)

**Relative velocity**

By ‘the velocity of object X relative to object Y’ we mean the velocity of X as measured by an observer moving with Y, who considers himself to be at rest.

Suppose that *we *observe X and Y and measure their
velocities relative to *us* to be v_{X} and v_{Y}. Then it can be shown that:

- velocity of X relative to Y = v
_{X}- v_{Y} - velocity of Y relative to X = v
_{Y}- v_{X}

**Example**

How long does it take for X to reach Y?

If we take velocities to the right as positive, then:
v_{X} = +12
ms^{-1} and v_{Y} = - 4 ms^{-1}.

The velocity of
X relative to Y = v_{X} -
v_{Y} = 12 -(-4) = 12 + 4 = 16
ms^{-1}.

Thus, if we treat Y as being at rest then X approaches Y at 16
ms^{-1}.

Each
trolley moves at its own original speed for this 2.5 s. Distance = speed * time,
so:

Trolley X moves 12*2.5 = 30m and trolley Y travels 4 * 2.5=10m, the
total being 40m, as expected.

**Relative acceleration**

If two objects X and Y have accelerations a_{X} and a_{Y}, then:

- the acceleration of X relative to Y = a
_{X}- a_{Y} - the acceleration of Y relative to X = a
_{Y}- a_{X}

Two object, X and Y, are at rest and 50m apart. At the same instant they start to accelerate as shown below:

- How long will it take for them to meet?
- How far from X’s original position will they then be?

The acceleration of X relative to Y = a

The equivalent diagram is:

**©
A **

A Level Physics - Copyright © A
C Haynes 1999 & 2004