1. Kinematics PH1.2

2. Motion – defining important terms:
(1) distance and displacement
scalar
vector
unit: m
displacement
distance
symbol: s

3. … the same speed, but different velocities
(2) speed and velocity
scalar
vector
… the same speed, but different velocities

4. distance time average speed = unit: m/s or ms-1 Δs t v = ds dt
from GCSE work …
or average velocity in a straight line
unit: m/s or ms-1
using symbols, we write …
v = Δs t
Δ means ‘a change in …’
If t is so small that there is not enough time for the speed to change …
d means ‘a very small change in …’ ds dt
instantaneous speed =

5. … is the rate of change of the velocity
(3) acceleration
not speed!!
… is the rate of change of the velocity
For motion in a straight line …
change in velocity
time taken
average acceleration =
a = Δv t
unit: m/s2 or ms-2
in symbols … dv dt
instantaneous acceleration =
in a very small time …

6. Motion graphs higher constant speed (1) Displacement-time graph
acceleration
displacement
constant speed
at rest
time
constant speed in reverse

7. speed = gradient of graph s
Calculating speed from a displacement-time graph
(1) constant speed
displacement
speed = gradient of graph s
time t
speed =
distance
time s t =

8. ds dt (2) instantaneous speed
displacement
speed is constant during a small time interval ds
enlarging a small part of the graph … dt
time

9. ds dt ds dt instantaneous speed = = gradient of this section
displacement ds dt
time

10. s instant. speed t = s t ds dt instantaneous speed =
= gradient of this line
displacement s t
instant. speed = s
the gradient is the same, irrespective of the size of the triangle t
time

11. change in instantaneous speed
Calculating acceleration from a displacement-time graph
change in instantaneous speed
time taken to change
acceleration =
displacement
(s2 / t2) - (s1 / t1)
T2 – T1
acceleration = T1 T2 t2 s2
time taken to change t1 s1
time

12. Motion graphs larger constant acceleration (2) Velocity-time graph
non-uniform acceleration
constant acceleration
velocity
constant speed
time
constant deceleration, then constant acceleration in reverse

13. acceleration = gradient of the graph
Calculating acceleration from a velocity-time graph
velocity
time v t
acceleration = gradient of the graph
acceleration =
change in velocity
time taken to change v t =

14. v t v = t dv dt instantaneous acceleration = = gradient of this line
velocity
instantaneous acceleration v t = v t
time

15. Calculating the distance travelled …
velocity v
time t
distance = speed × time = v·t
distance = area under the graph

16. Calculating the distance travelled during a constant acceleration …
velocity v
time t
distance = area under the graph = ½v·t

17. Calculating the distance travelled during a changing acceleration …
velocity
time
distance = area under the graph

18. distance = area under the graph
Calculating the distance travelled during a changing acceleration …
velocity
time
distance = area under the graph

19. distance = area under the graph distance = A1 + A2 + A3 + A4
Calculating the distance travelled during a changing acceleration … A2 A1 A3 A4
distance = area under the graph
distance = A1 + A2 + A3 + A4

20. The equations of motion
only considering motion in a straight line
constant acceleration only
we use symbols …
u – initial velocity
v – final velocity
s – displacement
t – time
a – acceleration

21. Equations of motion v = u + at By definition … change in velocity
time taken
acceleration =
or in symbols …
v – u t
a =
after rearranging …
v = u + at

22. average velocity = ½(u+v)
From the velocity-time graph for the journey …
velocity v
average velocity = ½(u+v) u
time t
distance travelled = average velocity × time
s = ½(u+v)t

23. From the first eqn. of motion, v = u + at
Substituting for v in the second eqn. of motion,
s = ½(u+v)t
s = ½(u +u + at)t
s = ½(2u + at)t
s = ½(2u + at)t
s = ½(2ut + at2)
s = ut +½at2
hence

24. v = u + at t = (v-u) / a 2as = uv + v2 – u2 – uv v2 = u2 + 2as
From the first eqn. of motion,
v = u + at so
t = (v-u) / a
Substituting for t in the second eqn. of motion …
s = ½(u+v)t
s = ½(u+v)(v-u) / a
s = (u+v)(v-u) / 2a
2as = (u+v)(v-u)
2as = uv + v2 – u2 – uv
2as = uv + v2 – u2 – uv
v2 = u2 + 2as
hence

25. Projectiles
a projectile is any object that moves solely under the influence of gravity
we ignore the effects of air resistance
we analyse motion in two dimensions

26. Dropping a projectile (i.e. freefall) …
Galileo’s experiment in Pisa, the hammer and feather experiment on the Moon
without air resistance, acceleration = 9.8ms-2 and is independent of the object’s mass

27. Launching a projectile horizontally …
a steel ball is pushed sideways
a steel ball is dropped
multi-flash photos show that …
vertical motion and horizontal motion are independent of each other

28. constant acceleration g = 9.8ms-2 downwards
Vertical motion
the distance between the lines is getting smaller
gravity pulling it downwards
constant acceleration g = 9.8ms-2 downwards
Horizontal motion
equal distance between lines
no force acting
constant speed sideways

29. Resolving a vector – splitting a vector into two components
concentrate on two perpendicular vectors θ v y x
vy = v·sinθ
vx = v·cosθ

30. Launching a projectile at an angle
uy = u·sinθ u
first of all, the initial velocity (u) needs to be resolved into components θ
vertical component of the velocity is changing throughout
ux = u·cosθ
horizontal component of the velocity is constant
the projectile’s speed and direction is constantly changing

31. time to climb and fall
= time available to travel sideways

32. force = mass × acceleration
driving force G
drag D
force = mass × acceleration
the acceleration is directly proportional to the magnitude of the resultant force
G - D = m × a
G - D
acceleration a = m

33. When the driving force > drag, the car accelerates
As the car accelerates, the drag increases
and the resultant force decreases
therefore the acceleration decreases,
but the car is still accelerating, so …
… this is repeated until driving force = drag
no resultant force → no acceleration → terminal velocity

34. during freefall, resultant force = weight – air resistance
On jumping, resultant force = mg and acceleration = g
When he starts to fall, air resistance appears
so the resultant force decreases and the acceleration < g …
but he is still accelerating and air resistance continues to grow…
so the resultant force and acceleration are still decreasing ...

35. during freefall, resultant force = weight – air resistance
This is repeated, until the speed is large enough for air resistance = parachutist’s weight ….
At this point, resultant force = 0 and acceleration = 0 ….
The parachustist reaches a constant speed, i.e. the terminal velocity

36. during freefall, resultant force = weight – air resistance
When the weight is small …..
only a small air resistance is needed to cancel the weight,
so a high speed is not needed for the air resistance to grow sufficiently…
therefore the terminal velocity is small