1. Deriving the equations
Kinematics
Deriving the equations

2. Objects moving under acceleration diagrams

3. Starter: covert these measures
KINEMATICS 1 KUS objectives BAT derive the suvat equations BAT apply them to solve problems BAT change between velocity measures
Starter: covert these measures
16 m/s to km/h
40 mph to km/h 3

4. Objects moving under acceleration
Imagine a particle moving in a straight line with constant (uniform) acceleration
What other measures should we consider in our model?
a acceleration
t time
Key Q: what do we mean by constant acceleration?; what is the difference between distance and displacement (s)
u initial velocity
v final velocity
s displacement
What are typical units for these measures?

5. First uniform acceleration equation
(1)

6. First uniform acceleration equation
WB1 A train starts from rest at a station and moves with constant acceleration. 35 seconds later it passes a signal box with a speed of 50 ms-1
What is its acceleration in ms-2 ? 𝑠=
u =
v =
a =
t =
50=0+𝑎(35)
𝒗=𝒖+𝒂𝒕
𝑎= = 𝑚𝑠 −2

7. Second uniform acceleration equation
(2)

8. Second uniform acceleration equation example
WB2 A boat accelerates from 4 ms-1 to 6.5 ms-1 over 8 seconds
How far has the boat gone in this time? 𝑠=
u =
v =
a =
t =
𝒔= 𝒖+𝒗 𝟐 𝒕 𝑠=
𝑠= 42 m

9. Third uniform acceleration equation
and
(3)
Key Q: Given first two eqns how can we use substitution to derive eqn 3 and 4?;

10. Third uniform acceleration equation example
WB A space shuttle travelling at 20 kms-1 slows with a deceleration of 45 ms-2 for 2 minutes. How many km will it have travelled? 𝑠=
u =
v =
a =
t =
𝒔=𝒖𝒕+ 𝟏 𝟐 𝒂 𝒕 𝟐
𝑠=20000×120+(−45) 𝑠=
𝑠=1752 𝑘𝑚

11. Fourth uniform acceleration equation
and so
(4)

12. Changing units of speed
Change these measures to metres per second or km per hour
20 km/h →
30 ms-1 →
Key Q: what steps to convert kmph to mph or ms-1
360 km/h →

13. Fourth uniform acceleration equation example
WB4 A car is following a Lorry going at 60 kmph on the motorway. The car constantly accelerates at 3 ms-2 to overtake the lorry and keeps accelerating for the 120 metres it takes to overtake. How fast is the car travelling once it has overtaken the lorry?
Give your answer in kmph. Does this seem likely for a real life model? 𝑠=
u =
v =
a =
t =
𝒗 𝟐 = 𝒖 𝟐 +𝟐𝒂𝒔
𝑣 2 = 60× (120)
𝑣 2 = ≅998
𝑣= × ≅114 𝑘𝑚𝑝ℎ
60 𝑘𝑚𝑝ℎ is 37.5 𝑚𝑝ℎ and 114 𝑘𝑚𝑝ℎ is 𝑚𝑝ℎ

14. Summary
If you know three of these values you can always use the equations to work out the other two

15. 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑠=20×10+(−4) 10 2 𝑠=−200 𝑚 𝑠= u = v = a = t =
WB 5 
An object decelerates uniformly at 4 ms-2 along a straight horizontal road. If the object travels for 10 seconds and starts at  20ms−1  
Find the displacement of the object
Describe the journey the object travels and find what distance it travelled
𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑠=
u =
v =
a =
t =
𝑠=20×10+(−4) 10 2
𝑠=−200 𝑚
The object slowed to a stop then travelled back, passing where it started and ending up 200m behind where it started

16. WB 6 
A motorcycle accelerates uniformly along a straight horizontal road so that, when it has travelled 30 metres, its velocity has increased from 15ms−1 to 18ms−1. 
Find the acceleration of the motorcycle
Find the time that it takes for the motorcycle to travel this distance. 𝑠=
u =
v =
a =
t =

17. 𝑠=𝒖𝑡+ 1 2 𝑎 𝑡 2 𝑠= 𝑣−𝑎𝑡 𝑡+ 1 2 𝑎 𝑡 2 𝑠=𝑣𝑡 −𝑎 𝑡 2 + 1 2 𝑎 𝑡 2
WB 7 The FIFTH equation 
There is a fifth equation of motion: 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2
Can you derive this from the other equations?
𝑠=𝒖𝑡+ 1 2 𝑎 𝑡 2
𝑠= 𝑣−𝑎𝑡 𝑡+ 1 2 𝑎 𝑡 2
𝑠=𝑣𝑡 −𝑎 𝑡 𝑎 𝑡 2
𝑠=𝑣𝑡 − 1 2 𝑎 𝑡 2

18. 𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2 160=28𝑡− 1 2 (2) 𝑡 2 s= 160 u = v = 28 a = 2
WB7 The FIFTH equation 
A cyclist reaches a straight stretch of road, of length 160 m, and travels its length with constant acceleration of 2 𝑚𝑠 −2 .
If the cyclist reaches the end of the straight stretch at velocity 28 𝑚𝑠 −1 , how long could it have taken?
𝑠=𝑣𝑡− 1 2 𝑎 𝑡 2
160=28𝑡− 1 2 (2) 𝑡 2
s= 160
u =
v = 28
a = 2
t =
𝑡 2 −28𝑡+160=0
(𝑡−20)(𝑡−8)=0
𝑡=20 seconds
or 𝑡= 8 seconds

19. 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 22=2𝑢+2𝑎 (1) 126=6𝑢+18𝑎 (2)
WB8 A car is moving along a straight horizontal road with constant acceleration.
There are three points A, B and C, in that order, on the road, where AB = 22m
and BC = 104 m
The car takes 2 secs to travel from A to B and 4 secs to travel from B to C
Find
The acceleration of the car
The speed of the car at the instant it passes A
From A to C
s= 126
u = u
v =
a = a
t = 6
From A to B
s= 22
u = u
v =
a = a
t = 2
𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2
𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2
22=2𝑢+2𝑎 (1)
126=6𝑢+18𝑎 (2)
Question from 2017 AS assessment materials
126=6(11−𝑎)+18𝑎
Combine (1) and (2)
126=66+12𝑎
𝑎=5 𝑚𝑠 −2
Substitute to (1)
𝑢=6 𝑚𝑠 −1

20. remember for next lesson ….
KINEMATICS 1 KUS objectives 1 derive the suvat equations 2 apply them to solve problems 3 change between velocity measures
One thing I will do
For the next lesson ….
One thing I need to
remember for next lesson ….
One thing I understood well is ….
Plenary pyramid

21. END