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As both stones fall, the difference in their velocities..increases decreases remains constant
As both stones fall, the difference in their heights (y-positions).. increases decreases remains constant
to determine rates of change, (or derivative) of a function, by calculating the slopeto determine the change of another quantity (or the integral) of a function, by calculating the area enclosed by the graph and the axis.
d=5.85 mvf=0.0 m/sAssume the skid distance, and assumed acceleration are accurate, and determine the initial speed given those assumptions.If the initial speed is greater than the speed limit, then can easily make a conclusionAssume they were travelling the speed limit and see what the skid distance would be if acceleration is 9.8 m/s2. We’d then check this result against 5.85 m. If the theoretical distance is longer than the 5.85 m, that would mean that the 5.85 m distance would be produced by someone who was travelling under the speed limit (and vice versa).Assume the skid distance, initial speed, and final speeds are correct, and determine what acceleration would be needed to stop the car. If the required acceleration is less than the assumed maximum value, then we can determine that they were travelling safely (and the opposite)
How far beyond the starting point will the car overtake the truck?What will the car’s speed be at the point it catches the truck? (in m/s and km/h)
As both stones fall, the difference in their velocities..
increases decreases remains constant
As both stones fall, the difference in their heights (y-positions)..
increases decreases remains constant
Equation 1 (and equation 5) are independent of aavg, Equation 2 is independent of d, Equation 3 is independent of vfEquation 4 is independent of vi
1-D Kinematics
Straight Line Kinematics
Interrogating Question
Two stones are dropped, one two seconds after the first, into a bottomless pit
Assume no air resistance.
Explain your choice:
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Explain your choice:
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Instruction
“All models are wrong, some are useful.”
- George Box
In grade 11, we spent a lot of time on uniform and non-uniform motion (alternative descriptions may have been constant velocity and constant acceleration kinematics).
Graphically
Uniform motion looks like:
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Non-uniform motion looks like:
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We can analyse the graphs:
Algebraically
Equation
d
aavg
vf
vi
t
Cannot be used to solve for…
v=aavgt
displacement
d=vavgt
acceleration
d=vit+12aavgt2
final velocity
d=vft-12aavgt2
initial velocity
vf2=vi2+2 a d
duration
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Worked Example 1
A police officer had to decide if a driver had been exceeding a 50.0 km/h speed limit before the driver made an emergency stop with brakes locked and tires sliding. The length of the skid was 5.85 m. The police officer made a reasonable assumption that the maximum braking acceleration of a car would not exceed the acceleration of a freely falling body. Was the driver charged with reckless driving?
We know two things for certain:
We can’t be certain about our answer, but can make a reasonable assumption.
Since most stated friction coefficients for tires on roads are 1.00.
This will result in a maximum acceleration of 9.8 m/s2 in a direction opposing motion.
We will then use this to determine a method of solving the problem.
All options are valid, you have to determine which one makes the most sense to you (you could use any of the other options to check if your answer is correct).
Given:
d=5.85 m vf=0.0 m/s a=-9.8 m/s2
Required:
vi=
Analysis:
vf2=vi2+2 a d
Solution
vi2=vf2-2 a d
vi=vf2-2 a d
vi=0.0 m/s2-2–9.8 m/s25.85 m
vi=10.7 m/s
vi=38.5 km/h
Statement:
The theoretical initial speed is 38.5 km/h. This is less than the stated speed limit of 50 km/h, so we can only say that it is entirely possible that the driver was driving under the speed limit before skidding.
Worked Example 2
At the instant a traffic light turns green, an automobile starts with an acceleration of 1.83 m/s2.
At that same instant a truck, traveling with a constant speed of 32.9 km/h, overtakes and passes the automobile.
Completion Problem 1
Completion Problem 2
Interrogating Question
Two stones are dropped, one two seconds after the first, into a bottomless pit Assume no air resistance.
Explain your choice:
There are no rows in this table
There are no rows in this table
Summarize
Developing Equations of Motion
From a d–t (position vs. time) graph of uniform motion (motion with constant velocity) we can use the slope to derive the first equation:
d=vavgt (1)
From a v–t (velocity vs. time) graph for motion with constant acceleration, we can use the slope to determine the acceleration.
aavg=vt
And rearrange that to get:
v=aavgt (2)
We can also look at this graph and use the fact that the area between the curve and the axis is the displacement (change in position) that the object undergoes.
We can determine the area in 3 ways:
Method 1
d=vit+12aavgt2 (3)
Method 2
d=vft-12aavgt2 (4)
Method 3
d=vi+vf2t (5)
At this point, we have 5 equations that do not include one variable.
What we don’t yet have is an equation that is independent of ∆t.
So, how do we do this?
We find two equations that describe the motion we are looking for and perform a substitution or elimination.
Working with 1 and 2 are probably easiest.
d=vavgt v=at
We can rearrange both to be equal to ∆t
(1) becomes t=dvavg, and (2) becomes t=va
If both of these equations are used in the same question and are interrogating the same time interval t1=t2, and so…
dvavg=va
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