3-kinematics.ppt general physics 1 PowerPoint

Physics 1 kinematics

1. © Boardworks Ltd 20051 of 37
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© Boardworks Ltd 20051 of 37
AS-Level Maths:
Mechanics 1
for Edexcel
M1.3 Kinematics
For more detailed instructions, see the Getting Started presentation.

2. © Boardworks Ltd 20052 of 37
C
o
n
t
e
n
t
s
© Boardworks Ltd 20052 of 37
Motion graphs
Motion graphs
Formulae for constant acceleration
Examination-style questions

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Kinematics involves the study of how things move.
It is only concerned with the motion itself, not the forces that
cause this motion.
The kinematics of an object is described in terms of its
Kinematics
distance,
displacement,
speed,
velocity,
acceleration.

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The distance a body has travelled is literally the amount of
‘ground’ it has covered during its motion.
Distance is a scalar quantity.
Displacement is a vector quantity.
Displacement describes how far a body is from its starting
point and in what direction.
Distance and displacement are measured in metres, m.
Distance and displacement

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The speed of a body relates to how fast the body is travelling.
Speed is a scalar quantity.
Velocity is a vector quantity.
The velocity of a body relates to how fast the body is travelling
and in what direction. It is the rate at which a body changes its
position.
Speed and velocity are measured in metres per second, ms
–1
.
Speed, velocity and acceleration
Acceleration can be a scalar or a vector quantity.
Acceleration is the rate of change of speed or velocity.
It is measured in metres per second per second, ms
–2
.
Negative acceleration is often called deceleration or
retardation.

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The kinematics of a body can be represented graphically.
The most common graphs are position-time, speed-time,
velocity-time and acceleration-time graphs.
Motion graphs
The gradient of a distance-time graph gives speed.
The gradient of a displacement-time graph gives velocity.
The gradient of a velocity-time graph gives acceleration.
The area under a speed-time graph gives the distance
travelled.
The area under an acceleration-time graph gives the change
in velocity.
The area under a velocity-time graph gives the change in
displacement.

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This graph shows a journey of 2000 m. It includes a stop of 1
hour after travelling 1000 m metres. The person then returns
to their starting position.
Displacement-time graph
The gradient of this graph
gives velocity.
For the first part of the journey,
= 500 metres per minute
0
Time (mins)
D
is
p
la
c
e
m
e
n
t

(
m
)
20406080100120
200
400
600
800
0
1000
For the second part of the
journey the velocity is zero.
velocity =
1000
20
For the last part, velocity =
1000
30

= –33.3 metres per minute

8. © Boardworks Ltd 20058 of 37
This graph also shows a journey of 2000 m with a 1 hour stop.
However, for this graph there is no indication of direction.
Distance-time graph
The gradient of this graph
gives speed.
For the first part of the journey,
= 500 metres per minute
0
Time (mins)
D
is
t
a
n
c
e

(
m
)
20406080100120
400
800
1200
1600
0
2000
For the second part of the
journey the speed is zero.
speed =
1000
20
For the last part, speed =
1000
30
= 33.3 metres per minute

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The gradient of the graph gives acceleration.
Velocity-time graph
The area under a velocity-time graph gives displacement.
0
Time (s)
V
e
lo
c
it
y

(
m
s
–
1
)
20406080100120
2.5
5
7.5
10
0
12.5
140
In this example, the area
under the graph is given by a
trapezium with height 12.5
and parallel sides of length
130 and 90.
Displacement =
1
2
(130+90)×12.5
= 1375 m
The first part of the graph shows an acceleration of 0.42 ms
–2
,
the second part 0 and the last part a deceleration of 1.25 ms
–2
.

10. © Boardworks Ltd 200510 of 37
This graph shows constant acceleration.
The first part shows an acceleration of 4 ms
–2
.
The second part shows a deceleration of 6 ms
–2
.
Acceleration-time graph
Time (s)
A
c
c
e
le
r
a
t
io
n

(
m
s
–
2
)
–2
0
2
4
–4
6
–6
2 121446 108
The last part shows constant velocity.

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The area under an acceleration-time graph gives change in
velocity.
Acceleration-time graph
For the first part, change in velocity = 4 × 8 =
For the second part, change in velocity = –6 × 3 =
32 ms
–1
–18 ms
–1
Time (s)
A
c
c
e
le
r
a
t
io
n

(
m
s
–
2
)
–2
0
2
4
–4
6
–6
2 121446 108
There is no change in velocity for the last part.

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Graphs example 1
A man travels in a lift from the top floor of a hotel to reception
on the ground floor.
b) sketch the acceleration-time graph of the lift.
It then travels at this constant velocity for t seconds before
decelerating with a constant deceleration of 2 ms
–2
until it
reaches the ground floor.
The lift accelerates with a constant acceleration of 1 ms
–2

until it reaches a constant velocity of 4 ms
–1
.
Given that the man has descended 44 m,
a) sketch the velocity-time graph of the lift and use it to find t

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Graphs solution 1
8 + 4 +4=44t
1 1
2 2
( ×4×4) + (4 ) +( ×4×4)=44t
The velocity-time graph for the lift can be sketched as follows:
The distance travelled is given
by the area under the graph, so
4 =32t
=8 secst
The acceleration-time
graph for the lift can
then be sketched as
follows:
Time (s)
A
c
c
e
le
r
a
t
io
n

(
m
s
–
2
)
–2
–1
0
1
2468101214
Time (s)
V
e
lo
c
it
y

(
m
s
–
1
)
4
4 t 2

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A car and a motorcycle are travelling along a straight road.
The car accelerates from rest to a constant speed of 28 ms
–1
.
The motorcycle accelerates from rest to a constant speed of
25 ms
–1
in 10 seconds.
After travelling for 90 seconds the car hits traffic and
decelerates to a constant speed of 22 ms
–1
in 5 seconds.
The motorcycle is unaffected by the traffic and maintains his
speed.
The motorcycle overtakes the car after they have both
travelled 3700 m.
Draw a speed-time graph and use it to find the time when the
motorcycle overtakes the car and how long the car was initially
accelerating for.
Graphs example 2

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Graphs solution 2
28
0
25
22
10 9095
Car
Motorbike
V
e
lo
c
it
y

(
m
s
–
1
)
Time (s)
Let t
1 be the time for which the motorbike is travelling with a
constant speed before it overtakes the car.
The motorbike overtakes the car after travelling 3700 m so,
1
12
( ×10×25)+( ×25)=3700t
1
125+25 =3700t
1
25 =3575t
1
=143 secondst
t
1

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So the motorbike accelerates for 10 seconds and then travels
at a constant speed for 143 seconds before overtaking the car.
Graphs solution 2
The motorbike travels for 153 seconds before overtaking the
car.
Let t
2
be the time for which the car is initially accelerating.
t
2
28
0
25
22
10 9095
Car
Motorbike
V
e
lo
c
it
y

(
m
s
–
1
)
Time (s)
153
This area represents 3700 m

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Graphs solution 2
Since the area under the graph for the car between 0 and 153
seconds is equal to 3700 so we can write,
1 1
2 22 2
( × ×28)+((90 )×28)+( (22+28)×5)+(58×22)=370 0t t 
14t
2 + 2520 – 28t
2 + 125 + 1276 = 3700
14t
2
= 221
t
2 = 15.8 (to 3 sf)
Therefore the car was initially accelerating for 15.8 s (to 3 s.f.)

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C
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Formulae for constant acceleration
Motion graphs
Formulae for constant acceleration
Examination-style questions

19. © Boardworks Ltd 200519 of 37
If a particle is moving in a straight line with a constant
acceleration then there are five equations of motion that can
be used to determine missing quantities.
Where
For vertical motion acceleration due to gravity is g, 9.8ms
–2
.
Formulae for constant acceleration
= +v u at
1
2
= ( + )s u v t
21
2
= +s ut at
21
2
=s vt at
2 2
= +2v u as
s = displacement in metres
u = initial velocity in ms
–1
v = final velocity in ms
–1
a = acceleration in ms
–2
t = time taken in seconds
These are sometimes called
the suvat formulae.

20. © Boardworks Ltd 200520 of 37
=
v u
a
t

v = u + at
The motion of an object with initial velocity u and final velocity
v over time t can be illustrated using a velocity-time graph.
By definition, acceleration is
the rate of change of velocity.
The constant acceleration a is
therefore given by the gradient
of the graph. So
at = v – u
v = u + at
Velocity (ms
–1
)
u
v
Time (s)t

21. © Boardworks Ltd 200521 of 37
s = ½(u + v)t
We can use the same graph to find the distance s travelled by
an object with initial velocity u and final velocity v over time t.
This distance is given by the
area under the graph.
This area is a trapezium with
parallel sides of length u and v
and width t. So
Velocity (ms
–1
)
u
v
Time (s)t
1
2
= ( + )s u v t
This can also be written as
+
=
2
u v
s t
 
 
 

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distance travelled = area of rectangle A + area of triangle B
s = ut + ½at
2
1
2
= + ( )ut t v u
=
v u
a
t

soat = v – u
This gives usdistance travelled =
1
2
+ ( )ut t at
21
2
= +s ut atSo

23. © Boardworks Ltd 200523 of 37
s = vt – ½at
2
distance travelled = area of rectangle C – area of triangle D
1
2
= ( )vt t v u 
We have shown that at = v – u
This gives usdistance travelled =
1
2
( )vt t at
21
2
=s vt atSo

24. © Boardworks Ltd 200524 of 37
v
2
= u
2
+ 2as
We can show that v
2
= u
2
+ 2as as using
v = u + at 1
+
=
2
u v
s t
 
 
 
2
1Rearranging equation to make t the subject gives
( )
=
v u
t
a

2Substituting this into equation
+
=
2
u v v u
s
a
  
  
  
2 =( + )( )as u v v u
2 2
2 =as v u
So
2 2
= +2v u as

25. © Boardworks Ltd 200525 of 37
A stone is thrown vertically upwards with a speed of 10 ms
–1

from a point 15 m above the ground. Find the maximum
height above the ground that the stone reaches and find the
time taken for the stone to reach the ground.
Taking  to be positive, the
information given in the question is:
u = 10
a = –g (Take g to be 9.8)
The question is asking for s when
v = 0, and for t when s = –15.
Constant acceleration example 1

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To calculate s when v = 0, given u and a requires the use of
v
2
= u
2
+ 2as.
Therefore, the maximum height above the ground that the
stone reaches is 20.1 m (to 3 s.f.).
Constant acceleration solution 1
0
2
= 10
2
+ 2(–9.8)(s) 
0 = 100 – 19.6s
19.6s = 100
 s = 5.10 (to 3 s.f.)

27. © Boardworks Ltd 200527 of 37
Therefore the stone reaches the ground after 3.05 s (to 3 s.f.).
Constant acceleration solution 1
To calculate t when s = –15, given u and a requires the use of
s = ut + at
2
.
1
2
–15 = 10t + (–9.8)t
2

1
2
–15 = 10t – 4.9t
2
4.9t
2
–10t – 15 = 0
Arranging all the terms on the left gives us the following
quadratic equation
t = 3.05 or t = –1.01 (to 3 s.f.)
Using gives the solution
2
4
2
b b ac
a
  

28. © Boardworks Ltd 200528 of 37
A particle moves in a horizontal line from a point A to a point
C, via point B. It has a constant acceleration of 1 ms
–2
and
passes point B after 6 seconds and point C after a further 4
seconds. Its velocity at C is 50 ms
–1
. Calculate the velocity at
A and the distances AB and BC.
Taking  to be positive the question firstly asks for u when
a = 1, t = 10 and v = 50. This requires the use of v = u + at.
Constant acceleration example 2
We can sketch the situation as follows
A B C
t = 6 t = 10
v = 50

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v = u + at
Therefore the particle passes A with a velocity of 40 ms
–1
.
Therefore AB is 258m.
Constant acceleration example 2
50 = u + (1)(10)
50 = u + 10
 u = 40
s = 240 + 18 = 258
The next part of the question asks for s when u = 40, a = 1 and
t = 6, requiring the use of s = ut + at
2
.
1
2
s = ut + at
2
1
2
s = 40(6) + (1)(6)
2
1
2

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Therefore AC is 450 m and so BC is 192 m.
Constant acceleration example 2
s = ut + at
2
1
2
The final part of the question asks for s when u = 40, a = 1
and t = 10, again requiring the use of s = ut + at
2
.
1
2
1
2s = 40(10) + (1)(10)
2
s = 400 + 50 = 450

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A ball falls off a cliff and lands on the beach 3.2 seconds later.
How high is the cliff?
Taking  to be positive the information given in the question
is
Therefore the height of the cliff is 50.2 m (to 3 s.f.)
Constant acceleration example 3
a = 9.8
t = 3.2
u = 0
s = ut + at
2
1
2
To calculate s requires the use of the formula s = ut + at
2
.
1
2
s = (0)(3.2) + (9.8)(3.2)
2
1
2
s = 50.2 (to 3 s.f )

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C
o
n
t
e
n
t
s
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Examination-style questions
Motion graphs
Formulae for constant acceleration
Examination-style questions

33. © Boardworks Ltd 200533 of 37
A ball is thrown vertically upwards with an initial velocity of u
ms
–1
from a point 1.2 m above the ground. It reaches a
maximum height of 23 m above the ground.
Calculate
a) the initial velocity
b) the velocity with which the ball strikes the ground
c) the total time the ball is in the air.
Examination-style question 1

34. © Boardworks Ltd 200534 of 37
a) Taking  as positive, v = 0, s = 21.8, a = –9.8
b) Method 1 - Using downward motion only
Solution 1
v
2
= u
2
+ 2as
0 = u
2
+ 2(-9.8)(21.8)
0 = u
2
– 427.28
u = ±427.28 (to 3 s.f.) initial velocity is 20.7ms
–1
(to 3 s.f.)
Taking  to be positive, u = 0, a = 9.8, s = 23
v
2
= u
2
+ 2as
v
2
= 0
2
+ 2(9.8)(23)
v
2
= 450.8
 The ball hits ground with a
velocity of 21.2 ms
–1
(to 3 s.f.)
v = ± 21.2 (to 3 s.f.)

35. © Boardworks Ltd 200535 of 37
b) Method 2 – Using whole motion
c) Taking  as positive, u = –427.28, v = 450.8, a = 9.8
Solution 1
Taking  to be positive, u = –427.28, a = 9.8, s = 1.2
v
2
= u
2
+ 2as
v
2
= 427.28 + 2(9.8)(1.2)
v
2
= 450.8
 ball strikes ground with a velocity
of 21.2 ms
-1
(to 3 s.f.)
v = u + at
 ball is in the air for 4.28 s (to 3 s.f.)t = 4.28 (to 3 s.f.)
v =  21.2 (to 3 s.f.)
( )
=
v u
t
a

( 450.8 427.28)
=
9.8
t


36. © Boardworks Ltd 200536 of 37
A car is travelling with a uniform acceleration along a
straight road.
It passes a point A with a velocity of 8 ms
–1
and 5 seconds
later it passes a point B with velocity 25 ms
–1
.
Find the velocity with which the car passes the mid-point of
AB.
Examination-style question 2
The distance AB needs to be found first: u = 8, v = 25, t = 5
s = 82.5
s = (8 + 33)(5)
1
2
s = (u + v)t
1
2
 the distance AB is 82.5 m and so the mid-point of AB is at a
distance of 41.25 m from A.

37. © Boardworks Ltd 200537 of 37
Acceleration now needs to be found: u = 8, v = 25, t = 5
The velocity of the car as it passes the mid-point of AB can
now be found: u = 8, a = 3.4, s = 41.25
Solution 2
a = 3.4
v = u + at
( )
=
v u
a
t


(25 8)
=
5
a

 the acceleration of the car is 3.4 ms
–2
.
 the car passes the mid-point of AB with
a velocity of 18.6 ms
–1
(to 3 s.f.)
v
2
= u
2
+ 2as
v
2
= 82 + 2(3.4)(41.25) = 344.5
v = ±344.5