(a) Describe the motion shown on the speed time graph.
(b) Calculate the acceleration for each part of the graph.
4
(c) Find the distance travelled in the first 4 seconds.
speed
(m/s)
3
2
1
0
1
2
3
4
5
6
Time (s)
7
8
9
Velocity
(a) Find the acceleration for each
part of the graph.
m/s
(b) Draw an acceleration time graph
3
(c) Find the maximum displacement
from the start.
2
(d) Find the final displacement.
1
0
-1
-2
-3
1
2
3
4
5
6
7
8
Time (s)
Sketch graphs
v
t
a
t
Ball falling from rest – up direction is positive
v
Notes:
+
0
t
-
All red lines have same gradient
– (on Earth this will be – 9.8 m/s2
as this is acceleration due to
gravity).
Above the time axis the ball is
moving upwards, below it is
moving downwards
In your group sketch a graph showing the motion of a ball
which is thrown up. Start the instant after the ball leaves your
hand. Take up as positive.
Now do tutorial questions 27 to 32
SAQ to Qu 24
2010 Higher paper Qu 1,2
Purple book Ex 1.2
Equations of motion
Third year
v=d÷t
no acceleration
Fourth year
a = (v – u) ÷ t
uniform acceleration
distance = area under speed time graph
Higher
v = u + at
uniform acceleration
s = ut + ½ at2
v2 = u2 + 2as
v = ½( u + v)
displacement = area under velocity time graph
Advanced Higher - accelerations which are not uniform
- very fast speeds, relativity
Deriving Equations of motion
v
v
t – time taken
u - initial velocity
v – final velocity
a – acceleration
s - displacement
u
Acceleration = gradient of graph
t
a=v–u
t
so
v = u + at
equation 1
t
Displacement = area under the graph
s = ut + ½(v – u)t
s = ut +½at2
but v = u + at so (v – u) = (u + at – u) = at
equation 2
Displacement = area under the graph
s = ut + ½(v – u)t = ut + ½vt -½ut
s = ½(u + v)
To eliminate t
v = u + at
so
t=(v–u)÷a
s=½(u+v)t
= ½ (u + v )( v – u ) ÷ a
2as = ( u + v ) ( v – u )
2as = uv – u2 + v2 – uv
2as = - u2 + v2
v2 = u2 + 2as equation 3
Note you are unlikely to be asked to derive this equation.
Examples
1. A car travelling at 20 m/s accelerates uniformly at 0.5 m/s2
until it is travelling at 30 m/s. Calculate the distance
travelled by the car during this time.
2. A toy rocket is launched vertically and reaches a height of
60 m. What was its launch speed?
Now try tutorial questions 33 to 36
Qu 37
a challenge, there is more than one way
to reach the same answer. Which do you find easier
Qu 39 to 42
Up to SAQ 36
Purple book Chp 1.3
Always check on
signs + - + - + -
Projectiles
The only force which acts on a projectile
is the force due to gravity ( weight)
We need to resolve the
velocity into its horizontal
and vertical components
v
v
Horizontal velocity
Vertical velocity
vH
vv
Down
+ ve
t
No force in horizontal
direction so constant
velocity
t
Weight acts downward so
accelerates at 9.8 m/s2
down
Example
A car travelling with a horizontal speed of 20 m/s goes off
the top of a cliff. It lands 30 m from the foot of the cliff
(i) How high was the cliff?
(ii) What was the car’s velocity just before it hit the
ground ?
Tutorial questions 43 to 46
SAQs up to 39
Purple book Ex 1.4
Extra question
satillite
Example
A basket ball player throws the ball at 600 to the horizontal
and scores a basket. The foot of the basket was 12m away. If
the ball takes 2s to reach the basket find:(a) The initial speed of the ball.
(b) The height of the basket above the initial position
of the ball.
Tutorial Qu 47 to 50
SAQ up to 41
Purple book Ex 1.5
Estimate your take off velocity in a standing long jump.
Step 1
Vertical jump
Measure maximum vertical displacement, sv
Calculate initial vertical velocity, uv and then the
time for jump, t.
Step 2
standing long jump
sh maximum horizontal distance
assume you stay in the air for the same length of
time as your vertical jump ie uv and t will be the same as step 1.
Calculate the horizontal velocity, vH
Step 3
calculate take off velocity from uv and uH
Do you think the assumption in step 2 is justified?
If not, is the calculated value for horizontal velocity too big or too
small?
The world record for the standing long jump is 3.71 m