1) Acceleration is a measure of how quickly velocity changes. It can represent a change in speed, direction, or both. 2) The slope of a velocity-time graph represents acceleration. The area under the graph represents displacement. 3) For objects experiencing uniform acceleration, displacement can be calculated using: s = ut + (1/2)at^2

1. LECTURE 4
Acceleration
IB Physics Power Points
Topic 2
Kinematics
www.pedagogics.ca

2. Defining Acceleration
Acceleration is used commonly to mean “speeding up”
Acceleration in physics has a very specific meaning.
Acceleration is a measure of how quickly velocity changes.
To change velocity an object must undergo an acceleration.
Recall – velocity is a vector. A change in velocity can be a change
in speed OR direction OR both speed and direction.
Example 1: a car travels around a corner bend in the road at a
constant speed of 45 km/h. Is the car’s speed changing?
No (read the question). Is the car’s velocity changing?
Yes, because the direction of travel is
changing. Therefore this car is also
accelerating.

3. Defining Acceleration
Example 2: The driver of a car travelling along a straight road
at 45 km/hr sees the light ahead turn red. The driver slows and
stops at the intersection. Is the car accelerating?
Yes. The car’s velocity is changing. In this case, the car is
slowing down. This would be a negative acceleration.
Mathematically acceleration can be calculated by:
2 1v vv
a
t t

 
 
In words, acceleration is the change in velocity over a time
interval DIVIDED by the time interval.
Unit for acceleration is ms-2 or m/s2 or m/s/s

4. Quick problem #1: A skater changes her velocity from 3.5 ms-1
[N] to 2.5 ms-1 [S] in 3.4 s. What is her acceleration?
Quick problem#2: A car travelling at 24 km/h speeds up at a
rate of 0.52 ms-2 for 3.7 s. What is the final speed?
2 1 2.5 3.5
3.4
v v
a
t
  
 

-2
-2
1.8 ms [ ]
= 1.8 ms [ ]
N
S
 
2 1v v a t  
2 6.67 (0.52)(3.7)v  
-1
2 8.6 ms or 31 km/hrv 

5. Velocity time graphs tell us how velocity changes. The slope
of a velocity time graph indicates acceleration.
t
v
[up]
This graph shows how the
velocity of a falling rock
changes. What would the
slope be?

6. Distance Travelled by Objects Moving at Constant Velocity
We can calculate the change in position of an object travelling at
constant speed very easily.
s v t   
Distance = velocity x time
DO NOT USE THIS EQUATION
FOR ACCELERATING OBJECTS!!
v
t
t
The graph shows a situation
where the above equation
applies. How would you
determine Δs from just the
graph?

7. Distance Travelled by Objects Moving at Constant Velocity
From this observation we can make a very important general
conclusion: THE AREA UNDER A VELOCITY-TIME GRAPH
REPRESENTS DISPLACEMENT (distance).
s v t   
v
tt
Δs = v x Δt would be the same
calculation as determining the
area under the graph for the
given time interval.
area v t  

8. Distance Travelled by Accelerating Objects
v2
Δt
For a uniformly accelerating
object the same principle can
be applied.
1
2
displacement area
area b h

 
2 2 1
1
Recall
2
So
s t v v v a t      
Initial
velocity
is zero
2
1
1
When
2
Substitution gives
s a t v zero   

9. Distance Travelled by Accelerating Objects
When there is an initial
velocity the graph looks
like this…
The total area under the curve
gives us the displacement. This
is the combined areas of the
triangle and the rectangle.
2
1
1
2
Mathematically
s v t a t    
v2
t
v1
21
2
s a t  
1s v t   
Remember this formula!

10. Distance Travelled by Accelerating Objects
Summary so far …….
1. The slope of a velocity vs time graph indicates
acceleration.
2. The area under a velocity vs time graph indicates
displacement.
3. To find displacement of an accelerating object we
can use the formula:
     2
1
1
2
s v t a t

11. Distance Travelled by Accelerating Objects
We can also use this graph
to derive a different
formula.
This time use a slightly different
calculation for the triangle:
 
    
 
    

  
1 2 1
2 1
1
1 2
1
( -v )
2
v
2 2
2
Mathematically
s v t t v
v t t
s v t
v v
s t
v2
t
v1
   2 1
1
( )
2
s t v v
1s v t   

12. Distance Travelled by Accelerating Objects
The critical equations for uniform acceleration are:
1.
2.
3.
We can use substitution to produce a final useful
equation.
4.

    
 
2 1
or
v vv
a v a t
t t
     2
1
1
2
s v t a t
 
  1 2
2
v v
s t
     
2 2
2 1 2v v a s

13. In your data booklet

14. Examples
A car travelling at 13.0 ms-1 coasts for 17 s decreasing its
velocity to 2.4 ms-1. How far did the car coast?
 
 

  

  
1 2
2
13 2.4
17 130 m
2
v v
s t
s
Identify variables. 1v2v t
Identify unknown  displacement.
Choose equation, substitute and solve:

15. Examples
What was the initial velocity of a rocket that
accelerates upwards at 0.175 ms-2 to reach a velocity
of 36.7 ms-1 over an altitude change of 1250 m?
   
 
  
  
 

2 2
2 1
2
1 2
2
1
-1
1
2
2
36.7 2(0.175)(1250)
30.2 ms
v v a s
v v a s
v
v
Hint: the equation used is very useful when you are
not given time.

16. Freefall
An object is said to be in freefall when the only force
acting on it is gravity. As we ignore air resistance in most
calculations, any falling object is in freefall. The
acceleration of a freefalling object depends on the force of
gravity. On Earth …
-2
9.81 ms [down]gravitya 
It is conventional in physics to use the letter “g” to
represent this special acceleration.
It is also conventional to use [up] as the positive
direction in the reference frame so in most cases we
substitute a = g = -9.81 ms-2.

17. Examples
A ball is tossed upwards with an initial velocity of 15.6
ms-1 from the roof of the school. The ball reaches
maximum height and falls back down, missing the edge of
the roof and hitting the ground below. The velocity at
impact on the ground was 26.5 ms-1 [down]. How long
was the ball in the air?
 
   
2 1
2 1 26.5 15.6
4.29 s
9.81
v v
a
t
v v
t
g



  
   
