# Kinematics - StowmarketPhysics

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14 Νοε 2013 (πριν από 4 χρόνια και 8 μήνες)

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Kinematics

Definitions

Displacement

Distance moved in a particular direction

Scalar

Quantity that only has magnitude (size)

eg speed, distance, temperature, pressure

Vector

Quantity that has magnitude (size) and direction

eg velocity, displacement, acceleration, force

Definitions

Examples

The aircraft flew due north at 300 ms
-
1

-

velocity

The aircraft flew 600km due north

displacement

The ship sailed SW for 200 miles
-

?

I averaged 7mph during the marathon
-

?

The snail crawled at 2 mms
-
1

along the bench
-

?

The sales rep’s round trip was 200km
-

?

displacement

speed

velocity

distance

Symbols and Units

Quantity

Symbol for quantity

Symbol for unit

displacement

s (or x)

m

time

t

s

velocity

v

ms
-
1

These symbols may be different to what you have
been used to at GCSE

beware!

Vector Questions

1.
A spider runs along two sides of a table. What is its
final displacement?

O

A

B

1.2 m

0.8 m

θ

Vector Questions

2.
You walk 3km due north, then 4km due east.

a)
What is the total distance you have travelled?

b)
Make a scale drawing of your walk, and use it to find your
final displacement. Remember to give both distance and
direction.

c)

3.
A boat leaves harbour and travels due north for a
distance of 3km and then due west for a distance of
8km. What is the final displacement of the boat with
respect to the harbour? The boat then travels a further
distance of 1km due south. What is the new
displacement with respect to the harbour?

Using the components of a vector

A vector quantity is a quantity with both magnitude (size) and
direction. Sometimes it can be helpful to find the component (or
effect) of a vector in a particular direction. In this activity, you will
gain further practice in calculating vector components, using
trigonometry, and with scale drawing.

Part 1: Working out the component of a vector in a particular
direction

You will look at two ways of doing this: by drawing and by calculation
(using trigonometry).

Using the components of a vector

Finding components by drawing

You are walking at 4.0 m s

1

in a direction 30
°

N of E. What is the
component of your velocity in an easterly direction?

Make a scale drawing of this on graph paper. The x
-
axis points east, the y
-
axis points north. You need to choose a suitable scale.

You need to find the component of velocity in the easterly direction. From
the end of the velocity arrow, draw a line straight line down to the x
-
axis, point A.

Measure from the origin O to A. Convert this distance using the scale into
ms

1
:

This method is limited by the precision with which you can draw and
measure. How precisely can you measure the angle? How precisely can
you draw the velocity arrow? How accurately can you measure the
length OA? All of these factors affect the precision of your final answer

Using the components of a vector

Finding components by calculation

You have a right
-
angled triangle OAB. You know the length of the long side
(the hypotenuse) and you need to find the length of one of the other
sides. They are related by the cosine of the angle AOB. Now:

hypotenuse

so you have:

cos 30
°

= OA = OA

OB 4.0 ms
-
1

Using the components of a vector

Re
-
arranging gives:

OA = 4.0 m s

1

×

cos 30
°

and calculation gives:

OA = 3.46 m s

1

This is the same answer as you found by scale drawing, but without the
uncertainties introduced by drawing.

Take care not to be misled by the apparent accuracy of this answer.

(The calculator gave 3.464 1016... m s

1.) If your speed is given as 4.0 m
s

1, you can only give the value of the component to two significant
figures, i.e. 3.5 m s

1.

Using the components of a vector

You are running at a velocity of 8 ms

1

in a north
-
easterly direction (i.e. at
45
°

to both N and E). Find the components of your velocity in an
easterly direction, and in a northerly direction, firstly by drawing, and
then by calculation using trigonometry.

Explain why these two components have the same magnitude.

As the velocity vector makes an angle of 45
°

with north, the component
north is

8 ms
-
1

x cos 45
°

= 5.7 ms
-
1

The component east can be calculated in the same way or by using the
45
°

angle between the vector and north to calculate the component
east from the sine function:

8 ms
-
1

x sin 45
°

= 5.7 ms
-
1

A boat is pulled along a canal using a rope tied to its bow.
The rope makes an angle of 15 degrees with the
centre line of the canal, and the force applied to the
rope is 1800N. Using maths or scale drawing, calculate
the force pulling the boat along the canal, and the
force pulling the boat to the side of the canal.

15
°

Fh

Fv

F
h

= cos 15 x 1800 N = 1739 N

F
v

= cos 75 x 1800 N = 466 N

Solving problems using components

You set off to run across an empty supermarket parking strip, 100 m
wide. You set off at 55
°

to the verge, heading towards the entrance
to the supermarket. Your speed is 8 m s

1
. How long will it take you
to reach the far entrance?

8

m

s

1
5
5
º
1
0
0

m
Exam Questions

Solving problems using components

A train is gradually travelling up a long gradient. The speed of the train
is 20 m s

1

and the slope makes an angle of 2
°

with the horizontal.
The summit is 200 m above the starting level. How long will it take
to reach the summit?

2
0
0

m
2
º
Solving problems using components

2
0
0

m
2
º

To find the time taken to travel to the top of the hill you need
first to find the distance along the slope. From the diagram in the
question, it is clear that sin2
°

= 200 / slope, so that the length of the
slope is:

slope = 200 / sin 2
°

= 5731m

Then the time taken is:

Time taken = distance / velocity = 5731m / 20 ms
-
1

= 4 mins 47 secs

Flying in a side wind

A bird flies at a steady speed of 3 ms

1

through the air. It is pointing in
the direction due north. However, there is a wind blowing from
west to east at a speed of 2 ms

1
.

1.
What is the velocity of the bird relative to the ground?

2.
What is the displacement of the bird, relative to its starting point,
after it has flown for 20 seconds?

3.
In what direction should the bird point if it is to travel in a
northerly direction?

Exam Question

1

a)
A boat’s speed through still water is 2ms
-
1
east across a river. The river runs north to south and is
20m wide. If the river flows south at 1ms
-
1
, how far
downstream does the boat reach the other shore?

b)
In which direction should the boat aim in order to get
straight across in the shortest possible distance?

2

A rope is used to pull a narrow boat along a canal. The rope is
pulled by a horse. The tension in the rope is 600N and
the rope makes an angle of 30
°

with the canal bank.
What force must be provided (by the rudder and keel) to
keep the boat travelling parallel to the bank?

Exam Question

1

a)
A boat’s speed through still water is 2ms
-
1
. It heads due east across a river.
The river runs north to south and is 20m wide. If the river flows south at
1ms
-
1
, how far downstream does the boat reach the other shore?

10m downstream (the boat travels 1m downstream for every
2m across

a)
In which direction should the boat aim in order to get straight across in the
shortest possible distance?

1 ms
-
1

2 ms
-
1

θ

sin
θ

= 1
÷

2

= 30
°

°

east
of north

Exam Question

b)
A rope is used to pull a narrow boat along a canal. The rope is pulled by a
horse. The tension in the rope is 600N and the rope makes an angle of 30
°

with the canal bank. What force must be provided (by the rudder and keel)
to keep the boat travelling parallel to the bank?

600 N

30
°

Force pulling the boat towards
the bank is opposed by the
rudder. So, resolve to find the
vertical component

= cos 60
°

x 600N

= 300 N

Question

A horse pulls a barge of mass 5000 kg along a canal using a rope 10m
long. The rope is attached to a point on the barge 2m from the
bank. As the barge starts to move, the tension in the rope is 500N.
Calculate the barge’s initial acceleration parallel to the bank.

F
h

= F x Cos
Φ

Sin
Φ

= 2m
/ 10m

Φ

=

11.5
°

F
h

= F x Cos
Φ

= 500N x Cos 11.5
°

= 489N

F

= ma

a

= F / m

= 489N / 5000kg

= 0.098 ms
-
2

5000kg

500N

2m

F
h

Φ

Homework

Page 13

Questions 1 to 4

1.
1270N [1] at angle 19.3
°

[1] to the horizontal [1]

2.
256 ms
-
1

[1] at 20.6
°

[1]

3.
a) [1]

b) 2.0 ms
-
1

[1] at 90
°

to the bank [1]

c) 32.5 s [1]

4.
954 N [1]