More graphs of motion
&
Kinematic equations
SEPTEMBER 11-14, 2015
Review: Slope of a P-T graph
We can calculate the average velocity from a P-T graph by
finding the slope of the graph.
Remember: vavg =
𝜟𝒙
𝜟𝒕
=
𝒙𝟏 −𝒙𝟐
𝒕𝟏−𝒕𝟐
= slope of P-T graph
v=
10 m – 0 m
1s –
0 s
Once we
know v,
we can
plot it on
an V-T
graph
= 10 m/s
Review: Slope of a P-T graph
What if it’s a curvy line?
We can still just pick two points on the graph and use the
𝜟𝒙
𝒙 −𝒙
formula: vavg = = 𝟏 𝟐
𝜟𝒕
𝒕𝟏−𝒕𝟐
v=
10 m – 0 m
2s –
1 s
Once we
know v,
we can
plot it on
an V-T
graph
= 10 m/s
Slope of a V-T graph
We can calculate the acceleration from a V-T graph by finding
the slope of the graph.
Remember: a =
𝜟𝒗
𝜟𝒕
=
𝒗𝟏 −𝒗𝟐
𝒕𝟏−𝒕𝟐
= slope of V-T graph
What do you think the slope of a V-T
graph tells us?
a=
10 m/s – 0 m/s
1s –
0 s
Once we
Acceleration!
know a,
we can
plot it on
an A-T
graph
= 10 m/s2
Area under Motion graphs
Another important thing to know is that the area under the curve
in a motion graph has meaning.
area under V-T graph = Δx
Δx = area under V-T graph
How do you find the area of a
triangle?
Area = ½ base * height
Area = Δx
Δx = ½ * 5 s * 50 m/s = 125 m
Test your understanding
Think about your answer, wait until I tell you, then show me the
answer by holding up the correct number of fingers.
.
a) Which line has the highest acceleration?
1
b) How do you know & how could you calculate
that acceleration?
# 1 has the steepest slope. We could calculate a
by finding the slope.
c) Which line has the greatest displacement?
# 2 has the greatest displacement.
c) How do you know & how could you calculate
that displacement?
# 2 has the largest area under the line. We could
calculate displacement by finding that area.
You do
Use the graph of the motion of a
toy train, below, to answer the
questions.
1) What is the acceleration between
points C and D?
2) How far does the train travel between
points A and C?
Kinematic Equations
Up until now, we’ve had just two kinematic equations:
vavg =
𝜟𝒙
𝜟𝒕
and a =
𝜟𝒗
𝜟𝒕
These can be re-written in to more user-friendly forms …
Let …
t = the time for which the body accelerates
a = acceleration
vi = the velocity at time t = 0, the initial velocity
vf = the velocity after time t, the final velocity
x = the displacement covered in time t
vf = vi + at
𝒗 𝒂𝒗𝒈 =
𝒗𝒊+𝒗𝒇
𝟐
x = vi t +
𝟏
𝟐
𝒂𝒕2
vf2 = vi2 + 2ax
Problem Solving Strategy

Diagram the problem

List known variables

Determine what you are trying to find.

Determine your strategy

Solve the problem

Evaluate your answer. Check whether the units, sign, and magnitude
make sense.
Acceleration Problems – We do
Grace is driving her sports car at 30 m/s when a ball rolls
out into the street in front of her. Grace slams on the brakes
and comes to a stop in 3.0 s. What was the acceleration of
Grace’s car?
Sketch:
Known variables:
What are we solving for:
Strategy:
Solution:
Evaluate answer:
Acceleration Problems – We do
Grace is driving her sports car at 30 m/s when a ball rolls out into the
street in front of her. Grace slams on the brakes and comes to a stop in
3.0 s. What was the acceleration of Grace’s car?
Sketch:
Known variables: Vi = 30m/s Vf = 0 m/s t = 3.0 s
What are we solving for: a
Strategy: Use vf = vi + at to solve for a.
Solution: a = (vf – vi)/ t
a = (0 m/s-30 m/s) / 3.0 s = -10 m/s2
Evaluate answer: Yes! Units are correct, sign makes sense, and
magnitude is reasonable.
Acceleration Problems – We do
What is the displacement after 10.0 s of a mass whose
initial velocity is 2.00 m/s and moves with acceleration
a = 4.00 m/s2 ?
Sketch:
Known variables:
What are we solving for:
Strategy:
Solution:
Evaluate answer:
Acceleration Problems – We do
What is the displacement after 10.0 s of a mass whose initial
velocity is 2.00 m/s and moves with acceleration a = 4.00 m/s2 ?
Sketch:
Known variables: t = 10.0 s, vi = 2.00 m/s, a = 4.00 m/s2
What are we solving for: x
Strategy: x = vi t + 𝟏 𝒂𝒕2
𝟐
Solution: 220 m
Evaluate answer: Yes! Distance and sign make sense.
Acceleration Problems – You do
1)
A car has an initial velocity of 5.0 m/s . When its
displacement increases by 20.0 m, its velocity becomes 7.0
m/s. What is the acceleration?
2)
The car accelerate from rest to 28 m/s in 9.0 s. What
distance does it travel?
3)
A Jet plane lands with a speed of 100 m/s and can
accelerate uniformly at a maximum rate of -5.0 m/s2 as it
comes to a rest. Can this plane land at an airport where
the runway is 0.80 km long?
Acceleration Problems – You do
1)
A car has an initial velocity of 5.0 m/s . When its displacement
increases by 20.0 m, its velocity becomes 7.0 m/s. What is the
acceleration?
2
2
Strategy – use vf = vi + 2ax
Answer: 0.6 m/s2
The car accelerates from rest to 28 m/s in 9.0 s. What distance
does it travel?
Strategy, find a then find x. Answer: 126 m
2) A Jet plane lands with a speed of 100 m/s and can accelerate
uniformly at a maximum rate of -5.0 m/s2 as it comes to a rest.
Can this plane land at an airport where the runway is 0.80 km
long? Usevf2 = vi2 + 2ax
Answer: No – it needs 1000 m
2)