AVERAGE ACCELERATION

average acceleration = change in velocity / change in time.  (a_av = dv / dt)
Whenever velocity is changing - in either magnitude or direction - the motion is said to be accelerated.
In a car, what are you using to change the velocity ? (3 things)

Your car can get from 0 to 60mi/hr in 9.2 s.
The average acceleration is __________mi/hr/s
The speed increases by ____________mi/hr   in each second of the trip/

Initial velocity will be noted Vo (for V at t=0s) or Vi (for V initial). V1 is another equivalent notation.
the acceleration will be noted a. you will see that a can be negative if you are slowing down. (but moving to the right)
The final velocity will be noted V or sometimes V2 or Vf (for V final)
t is the time elapsed during the motion we are studying.
d is the displacement. d can be negative. 5 blocks to the right + 8 block to the left = displacement of -3
tells you, at the end of the motion, how far you are from the origin.

1) Suppose a car is traveling at 12 m/s and increases its speed uniformly to 30m/s, taking 15 s
to do so. Uniformly means the acceleration is constant. What is the value of the acceleration ?

2) What is the acceleration of a rocket in outer space that takes 5.0s to increases its speed from 1,240 m/s to 1,300 m/s ?

3) How much does the speed of a car increases if it accelerates uniformly at 2.5 m/s² for 5s ?

4) A car is going 8.0 m/s on an access road in a highway, and then accelerates at 1.8 m/s² for 7.2 s.
How fast is it then going ?

5) A plane is taking off. at t= 0s Vi =0 and and t = 29s v = + 260km/h
Find the average acceleration in km/h/s.


ACCELERATION IS A VECTOR

Observe the following picture:

What is going on ?
If V1 = + 10 m/s  and V2 = + 30 m/s
Find DV = V2 - V1 = ______ (don't forget the sign, it shows the direction)
a = DV/DT  a is a vector. a It has the same direction than DV
If the car changes its velocity in 5 s find a = ________  (don't forget the sign)

answers

Observe the following situation:


What is going on ? Predict the sign of V2 - V1
If V1 = + 30m/s (the direction East is represented by a positive sign )
V2 = + 10 m/s
Find V2 - V1 =  ________  (don't forget the sign. can be negative)
if DT = 2s (time elapsed between the 2 situations)
Find a = _________  !! a is a vector. Use the + or - signs to show the direction.
which way points a ? Can you trace a on the dia gram ?

SO:
If a car or an object is slowing down,  the vector velocity and the vector acceleration have opposite direction and therefore opposite signs.
So if  an object is moving to the right and is slowing down,  velocity has a positive sign (moving to the right)
but  acceleration has a negative sign and is pointing to the left.

V2 < V1

6) A drag racer crosses the finish line, and the driver applies the brakes to slow down.
The brakes are initially applied when to=9.0s and the car's velocity is + 28m/s.
When t = 12.0 s, the velocity has been reduced to v = +13m/s.
a) What is the average acceleration of the dragster ? (acceleration is a vector, it can be + or -)
b) Can you trace the vectors (V1, V2 and a) ? scale : 2 cm = 10 m/s (use a graph paper)


7) The driver of a car steps on the brakes, and the velocity drops from 20m/s due east in a time
of 2.0 s. WHat is the acceleration ? (acceleration is a vector. show the direction)
a = _____ @ _______

8) A runner moving with an initial velocity of 9.0 m/s slows down at a constant rate of - 1.5 m/s/s
over a period of 2 seconds. What is her velocity at the end of this time  ? after 1 second ?
(the acceleration can be negative (you are slowing down) but the velocity be positive (you are still
moving to the right even if slower) hints: use V = Vo + at but use integers (a <0 and Vo >0 )

9) Starting from rest and moving in a straight line, a runner achieves a velocity of 7 m/s in a time of 2s.
What is the average acceleration of the runner ? (if you are speeding up the acceleration is positive)
hints: Vo = 0 and Vf = + 7m/s

INSTANTANEOUS ACCELERATION

Is the rate of change of velocity at given time.
It is computed by finding the average acceleration for a very short
time interval during which the acceleration does not change appreciably.
Here we will only deal with a constant acceleration.

You stomach is our detector of instantaneous acceleration: It will senses the downward acceleration of an elevator.


GRAPHING MOTION AND THE BIG FIVE

try without hints. unless you get "confused"
10) The table gives some data for a Ferrari racing car at the start of a Grand Prix race:

speed ( m/s)	0	10	20	30	40	50	60
time (s)	0	1	2	3	4	5	6
distance (m)

A) Plot a speed-time graph for the car (x = time and y = speed)
B) Find the slope. What does the slope represent ? pick one : speed ? acceleration? displacement?
C) The acceleration of the car a = __________
C) Can you find the distance covered by the car at a given time ? complete the table above.
 remember distance = area under the curve speed vs time
hints: for example. for t =2 area = 0.5 (speed) (time) = 0.5 (20) ( 2) = 20m

D) so at a given time t and given the velocity V,
you can find the distance covered by the car using the formula:

d = _______  

hint: area = 0.5 (base) ( height) with base = t and height = V
but V at a given time t  is  : V = a t  (where a is the acceleration)
so d = ____________ use a and t to write the equation of motion.
hint: you should find d = 0.5 a t² . make sure you understand why. some algebra is involved.

E) Let's find the average velocity of the car 2 different ways:
i) average velocity = total distance / total time.   so find the total distance and the total time
hint: the total time is 6s and you found the total distance d for that time  in D)
remember ? d = area of the triangle of base 6 and height 60.
ii) Look at your graph. Can you find the average velocity by observing the graph? or the table ?
hint: the average is in the middle of all the values. this is because the acceleration is constant.

The previous problem has shown you:
Vav = (Vi + Vf) /2    where Vi is the initial speed (0 m/s) and Vf is the final speed (60m/s)
d = 0.5 a t² when Vi = 0


11) Suppose now that the car has an initial speed of 10m/s at t=0s  Vo= 10 m/s
A)Can you fill the table again. Keep the same acceleration. (10m/s²)

speed ( m/s)	10
time (s)	0	1	2	3	4	5	6
distance (m)

B) plot the graph speed vs time.
C) Can you find the distance covered at any time by computing the area under the line?
hint:the area under the line = area of the triangle + area of the rectangle. Fill the table.
example:for  t = 1s , distance d = 0.5 (1) ( 10) + (1)(10) =5 + 10 =15m, for t= 2s you should find d = 40m ...
the area under the line  = area of a triangle + area of a rectangle.
D) You just have found another equation of motion:
d = area of the rectangle +  area of the triangle
 and because V = Vi + at and using algebra you get: d = Vit + 05 at²
CHeck the displacement you found by using the formula : d = Vit + 0.5 at²

D) What is the AVERAGE speed ?
hint: try average speed = total distance / total time
you can also look at the speed: 10, 20, 30 , 40 , 50 , 60 , 70
and find the value in the middle. Or you can do :( Vi + Vf ) /2 (take the average between the first and the last speed)

E) or can also find the average speed by finding the middle between
the initial speed Vo (10m/s) and the final speed (70m/s)
Find (Vo + Vf)/2. Do you find the same value ? _________

THe graph should have looked like this one:



The displacement is the area under the graph velocity versus time.
You can break the area into 2 parts. A rectangle (displacement covered if the car does not
accelerate but keeps the same velocity of 10 m/s) and a triangle (the additional displacement due
to the acceleration. More distance is covered because the car accelerates).


THE BIG FIVE (write them on index cards)
Initial velocity will be noted Vo (for V at t=0s) or Vi (for V initial). V1 is another equivalent notation.
the acceleration will be noted a. you will see that a can be negative if you are slowing down. (but moving to the right)
The final velocity will be noted V or sometimes V2 or Vf (for V final)
t is the time elapsed during the motion we are studying.
d is the displacement. d can be negative. 5 blocks to the right + 8 block to the left = displacement of -3
tells you, at the end of the motion, how far you are from the origin.

So far you know:
average velocity = total displacement / time elapsed  or Vav = d / t
Both are vectors. !! If the velocity is not a constant, then the velocity V, at a given time, is not equal to the average velocity. (only true if there is no acceleration)
Displacement can be negative or positive and so can the average velocity.
Vav = (Vi + Vf ) /2          
Vav is the average velocity, Vi is the initial velocity (t= 0s) and Vf is the final velocity.
 In that case, the acceleration is non zero. We suppose the acceleration  constant.
a = (Vf - Vi) /t     or acceleration = change in velocity / time elapsed                 
We suppose the acceleration  constant. Otherwise, this becomes the equation for the average acceleration.
V = Vi + at
THe final velocity V, reached after a time t, is given by this equation. Vi is the initial velocity (t = 0s)
We suppose the acceleration constant.
d = Vi t + at² / 2 
d is the vector displacement. d can be negative or positive and so can be a or Vi (vectors too)
This equation give us the displacement after a time t. Vi is the initial velocity. a is the acceleration.
We suppose a constant. 
If Vi = 0 then d = at² / 2  also   if  a = 0  then d = Vi t  = V t    the velocity stays constant.

Playing with the equations:
d = Vi t + at²/2      
a = (Vf - Vi )/t
 d = Vav t with Vav = (Vi + Vf) /2
you can find:

Vf² = Vi² + 2ad                   remember if you move due west d <0. if you are slowing down a < 0

The 5  fundamental quantities used to decribe an uniformly accelerated motion along a straight line are:
Vi, Vf , ___ , _____ , _____

These five quantities are related by a group of five equations that we call the big five/
They work in cases where acceleration is constant.

Can you write these five equations ? (see above)

		variable that's missing
five 1	(displacement, a vector) d =	a acceleration
five 2	(velocity, a vector) V =	d displacement
five 3	d =	V (Vf)  final speed
five 4	d =	Vo initial speed
five 5	(final speed squared) V2 =	T time elapsed

answers

12) The velocity of  an automobile changes over an 8.0 s time period as shown
in the following table:

time (s)	velocity (m/s)	time (s)	velocity (m/s)
0.0	0	5	20
1	4	6	20
2	8	7	20
3	12	8	20
4	16

A) plot the velocity-time graph of the motion. Use a graph paper please.

B) Determine the displacement of the car during the first 2.0 s ?
hint: Find the area of the triangle with base = 2s  and height = 8m/s

C) What displacement does the car have during the 4.0 s?
hint: Find the area of the triangle with base = 4s  and height = 16m/s

D) What displacement does the car have during the entire 8.0 s?
displacement = area of the triangle + area of a rectangle.
hint: base of the triangle = 5 s and height = 20m/s
for the rectangle : base =3s, height = 30m/s
E) Find the slope of the line between t= 0s and t = 4s. WHat does this slope represent ?
(hint : the units will help you)

F) Find the slope of the line between t=5.0s and t = 7.0s . What does this slope indicate ?

MORE PROBLEMS:

13) a jet is taking off from the deck of an aircraft carrier. starting from rest (Vi = 0m/s) , the jet is catapulted
with a constant acceleration of + 31 m/s² along a straight line and reaches a velocity (Vf) of 62 m/s.
Find the displacement of the jet.
hint : first find the time using a = (Vf - Vi) /t
Then find the displacement using the formula you found previously d = Vi t + 0.5 a t²

14) a speedboat has a constant acceleration of 2 m/s² . If the initial velocity of the boat is 6 m/s,
calculate its displacement after 8.0s

15) How long does it take a car, starting from rest, to travel 240m if it its acceleration is 1.90 m/s² ?

16) What is the acceleration of the rocket that accelerates uniformly from rest and travels
650m in the first 12 s ?

17) An object with an initial velocity of 4 m/s moves along a straight  line under constant acceleration.
3 seconds later, its velocity is 14 m/s .   How far did it travel during this time ?


18) A car's that's initially traveling at 10m/s accelerates uniformly  for 4 seconds at a rate of 2 m/s²
in straight line/ How far the car travel during this time ?




Try:
19) A car going 22 m/s (initial velocity )  has its brakes jammed on and leaves skid marks 45m long.
what was its acceleration ? (the car stops) hints: Vi = 22m/s and Vf = 0 m/s  the acceleration is negative

You could have taken a short cut and use the formula:
Vf² = Vi² + 2ad   (you can derive this by using:  Vf= Vi + at and   d = Vi t + 0.5 a t²  )
20)
How fast is a car going if it starts at rest and accelerates uniformly at 2.8 m/s² while traveling 220 m ?







QUIZ or homework

LABS

0) accelerated motion

1) simulation: acceleration of a car.
Using the equations and given initial conditions (Vo )
predict the distance covered in T = 10s and the time elapsed when X = 30m

2) materials : incline planes, stop watch, steel ball, block, masking tape, pencil ..
print out:
part1     part2    

3) materials: protractor, string, small mass
part 1    part2

4) review graphs :

displacement  vs time
open this WINDOW
Play the animation 1 and try to trace  the graph : distance versus time/ Don't look at the graph yet.
When you are done, check with the view graph 1 key.
Do the same thing with the following animations.

velocity vs time
open this WINDOW
same thing

acceleration versus time
open this WINDOW

5) Watch this video (U-tube)   WINDOW
it was race between Charles Barkley and Dick bavetta.
Make sure you note the distance covered by the racer and the time.
Some conversion will be needed.
Find the acceleration of the winner and his final speed.
We suppose the acceleration is constant. ( Like if it were a race along a straight line).
What are the sources of errors ?
no picking.
answer