KINEMATICS LESSON 2 PART 2
WATCH ANIMATION

Passing Lane - Position vs. Time Graph
Observe two cars.
The blue car starts before the red car but the faster red car passes the blue car.
Observe the position-time graphs for these two cars. The slope of the lines indicate they
both have a constant, but different velocity.
At the time that the cars are side by side, the lines intersect. That is, the two cars share the
same position at that instant in time.
Note: These two cars start at different times and at different locations. For our purposes,
do the following question as if they both start at 4 seconds into the race, the blue car with
a 40 m head start.
Example: The Blue car starts 40 meters before the red car (this is why you don’t see the
blue car and red car start together) The blue car has a velocity of 6.0 m/s and the red car
has a velocity of 20.0 m/s. Where and when will the red car catch up with the blue car?
Draw a position-time graph, using the slopes of the lines, you can roughly determine
when the two cars will pass.
dblue = dred at catch up point
dblue = vbluet + do How come you changed this equation and not the other?
dred = vredt
therefore when red catches blue, dblue = dred
vbluet + do = vredt
6.0*t + 40 = 20.0*t
t = 2.8 seconds after the red car begins race (Blue car is already 40 m ahead)
Substitute t into one of the equations to find the distance when they pass:
dred = vredt = 20.0*2.8 = 56 m
distance when passing each other = 56 m
Example: A train leaves the station traveling North at 50.0 km/h. Leaving at the same
instant, a train leaves the station traveling South at 65 km/h. The distance between the
two train stations is 120 km. When and where do they meet?
Recall, that when two velocities are opposite directions they have to have opposite signs,
therefore we will call the velocity going South, negative (-65 km/h).
Draw a position-time graph, using the slopes of the lines, you can roughly determine
when the two trains will pass.
dnorth = dsouth at catch up points
Since the train leaving the South is designated as having a negative velocity, it starts at
120 km point and the train leaving the North starts at 0 (as the slope of the velocity is
positive)
dsouth = vsoutht + do
dnorth = vnortht
vsoutht + do = vnortht
-65t + 120 = 50t
t = 1.043 = 1.0 hour
Substitute t into one of the equations to find the distance when they pass:
dsouth = vsoutht + do = -65(1.043) + 120 = 52.17 = 52 km from the North
Why is it from the North? (look to graph for answer, what does d represent)
Describing Motion with Velocity vs. Time Graphs
The Meaning of Shape for a v-t Graph
***What if a line crosses over the x-axis from the positive region to the negative region
of the graph (or vice versa)?
then the object has changed directions.
See Animations (rightward velocity with rightward acceleration  leftward velocity with
rightward acceleration)
The Importance of Slope
The slope of the line on a velocity-time graph is the acceleration of the object.
Consider a car moving with a constant velocity of +10 m/s.
What is its acceleration? 0 m/s2
Now consider a car moving with a changing velocity.
Time
Velocity
(s)
(m/s)
0
1
2
3
4
5
0
10
20
30
40
50
AREA UNDER A v-t GRAPH
Area under a v-t graph, between any two times, is the displacement
Area = length x width
d =v
x t
Example: Determine the displacement of the object during the first second (Practice A)
and during the first 3 seconds (Practice B).
Example: Determine the displacement of the object during the time interval from 2 to 3
seconds (Practice A) and during the first 2 seconds (Practice B).
ANIMATION - Stop Light
-make observations and answer the questions
HOMEWORK LESSON 2 PART 2
1. A parent walking at 1.5 m/s is already 30 m down a walking path when a child begins
walking at 3.0 m/s to catch up. When and where does the child catch up?
Draw a position-time graph, using the slopes of the lines, you can roughly determine
when the two will pass.
dparent = dchild at catch up point
dparent = vparentt + do
dchild = vchildt
therefore when the child catches up with the parent, dparent = dchild
vparentt + do = vchildt
1.5*t + 30 = 3.0*t
t = 20 seconds
Substitute t into one of the equations to find the distance when they pass:
dchild = vchildt = 3.0*20 = 60 m
2. A truck travels at 80 km/h. A car traveling at 120 km/h departs 0.50 h later than the
truck. When and where does the car catch the truck
Draw a position-time graph, using the slopes of the lines, you can roughly determine
when the two will pass.
The equations must be made equal, therefore which ever equation does not begin at the
origin (0,0), must be altered to take the difference into consideration.
dtruck = dcar at catch up point
dcar = vcar(t - to)
dtruck = vtruckt
therefore when the car catches up with the truck, dtruck = dcar
vtruckt = vcar(t - to)
80*t = 120*(t – 0.5)
80t = 120t – 60
t = 1.5 hours
Substitute t into one of the equations to find the distance when they pass:
dtruck = vtruckt = 80*1.5 = 120 km
3. A car A leaves Penticton traveling north at 80 km/h. Car B leaves Vernon 90 km
away traveling at -100 km/h south. When and where do they meet. Remember to make
it directional and the one with the negative velocity must have a negative slope, therefore
starting at the 90 km mark.
Draw a position-time graph, using the slopes of the lines, you can roughly determine
when the two will pass.
dvernon = dpenticton at catch up points
Designated the car leaving the South as having a negative velocity, it starts at 90 km
point and the car leaving the North starts at 0 (as the slope of the velocity is positive)
dvernon = vvernont + do
dpenticton = vpentictont
vvernont + do = vpentictont
-100t + 90 = 80t
t = 0.5 hours
Substitute t into one of the equations to find the distance when they pass:
dpenticton = vprnyictont = 80*0.5 = 40 km from the South (Penticton)
4. Runner A gives B a 20 m head start head start in a race. A runs at 5.0 m/s and B at 3.0
m/s. Where and when will A catch B.
Draw a position-time graph, using the slopes of the lines, you can roughly determine
when the two will pass.
dA = dB at catch up point
dB = vBt + do
dA = vAt
therefore when the child catches up with the parent, dA= dB
vBt + do = vAt
3.0*t + 20 = 5.0*t
t = 10 seconds
Substitute t into one of the equations to find the distance when they pass:
dA = vAt = 5.0*10 = 50 m
5. Slow sam begins hiking at 12:00 at 2.5 km/h. Quick Quentin leaves 1.5 h later hiking
at 4 km/h. When and where does he catch Sam?
Draw a position-time graph, using the slopes of the lines, you can roughly determine
when the two will pass.
The equations must be made equal, therefore which ever equation does not begin at the
origin (0,0), must be altered to take the difference into consideration.
dsam = dQuen at catch up point
dquen = vquen (t - to)
dsam = vsamt
therefore when the car catches up with the truck, dsam= dquen
vsamt = vquen (t - to)
2.5*t = 4*(t – 1.5)
2.5t = 4t – 6
t = 4 hours
Substitute t into one of the equations to find the distance when they pass:
dsam= vsamt = 2.5*4 = 10 km
6. Student X jogs down a hallway at 5 m/s. Leaving at the same instant from the
opposite end of the 40 m long hallway Student Y walks quickly toward X at –2 m/s.
When and where do they meet?
Draw a position-time graph, using the slopes of the lines, you can roughly determine
when the two will pass.
dY = dXat catch up points
Since Student Y has a negative velocity, the slope of its line on the position time graph
must be negative, therefore student Y starts at position 40 m and student X at 0 m.
dY = vYt + do
dX= vXt
vYt + do = vXt
-2t + 40 = 5t
t = 5.7 s
Substitute t into one of the equations to find the distance when they pass:
dX = vXt = 5*5.7 = 28.6 m from student X
7. Consider the graph at the right. The object whose motion is represented by this graph
is ... (include all that are true):
a. moving in the positive direction.
b. moving with a constant velocity.
c. moving with a negative velocity.
d. slowing down.
e. changing directions.
f. speeding up.
g. moving with a positive acceleration.
h. moving with a constant acceleration.
Answer: Answers: a, d and h apply.
a: TRUE since the line is in the positive region of the graph.
b. FALSE since there is an acceleration (i.e., a changing velocity).
c. FALSE since a negative velocity would be a line in the negative region (i.e., below the
horizontal axis).
d. TRUE since the line is approaching the 0-velocity level (the x-axis).
e. FALSE since the line never crosses the axis.
f. FALSE since the line is not moving away from x-axis.
g. FALSE since the line has a negative or downward slope.
h. TRUE since the line is straight (i.e, has a constant slope).
8. Describe the motion depicted by the following velocity-time graphs. In your
descriptions, make reference to the direction of motion (+ or - direction), the velocity and
acceleration and any changes in speed (speeding up or slowing down) during the various
time intervals (e.g., intervals A, B, and C).
a.
Answer: The object moves in the + direction at a constant speed - zero acceleration
(interval A). The object then continues in the + direction while slowing down with a
negative acceleration (interval B). Finally, the object moves at a constant speed in the +
direction, slower than before (interval C).
b.
Answer: The object moves in the + direction while slowing down; this involves a
negative acceleration (interval A). It then remains at rest (interval B). The object then
moves in the - direction while speeding up; this also involves a negative acceleration
(interval C).
c.
Answer: The object moves in the + direction with a constant velocity and zero
acceleration (interval A). The object then slows down while moving in the + direction
(i.e., it has a negative acceleration) until it finally reaches a 0 velocity (stops) (interval B).
Then the object moves in the - direction while speeding up; this corresponds to a acceleration (interval C).
9. Determine the acceleration of the object depicted in the velocity-time graph at
intervals 0-4 seconds and 4-8 seconds
Time
Velocity
(s)
(m/s)
0
1
2
3
4
5
6
7
8
2
2
2
2
2
4
6
8
10
Answer: Find the slope of the lines, 0 4 s = 0 m/s2, 4 8 s = 2 m/s2
10. The velocity-time graph for a two-stage rocket is shown below. Use the graph and
your understanding of slope calculations to determine the acceleration and displacement
of the rocket during the listed time intervals.
a. t = 0 - 1 second
b. t = 1 - 4 second
c. t = 4 - 12 second
Answer: a. The acceleration is +40 m/s/s. The acceleration is found from a slope
calculation. The line rises +40 m/s for every 1 second of run. Displacement = 40 m/s x
1.0 s = 40 m
b. The acceleration is +20 m/s/s. The acceleration is found from a slope calculation.
The line rises +60 m/s for 3 seconds of run. The rise/run ratio is +20 m/s/s.
Displacement = 40 x 3 + ½ x 60 x 3 = 120 +90 =210 m
c. The acceleration is -20 m/s/s. The acceleration is found from a slope calculation.
The line rises -160 m/s for 8 seconds of run. The rise/run ratio is -20 m/s/s.
Displacement = ½ x 5 x 100 + ½ x 3 x –60 = 250 – 90 = 160 m remember, when
an object has a positive and then negative velocity, it is moving in opposite
directions, therefore the distances are subtracted
11. When lines on a velocity-time graph intersect, does it mean that the two cars are
passing by each other?
Answer: No, it means they have obtained the same velocity at that point in time.
HOMEWORK LESSON 2 PART 2
1. A parent walking at 1.5 m/s is already 30 m down a walking path when a child begins
walking at 3.0 m/s to catch up. When and where does the child catch up?
Draw a position-time graph, using the slopes of the lines, you can roughly determine
when the two will pass.
2. A truck travels at 80 km/h. A car traveling at 120 km/h departs 0.50 h later than the
truck. When and where does the car catch the truck
Draw a position-time graph, using the slopes of the lines, you can roughly determine
when the two will pass.
The equations must be made equal, therefore which ever equation does not begin at the
origin (0,0), must be altered to take the difference into consideration.
3. A car A leaves Penticton traveling north at 80 km/h. Car B leaves Vernon 90 km
away traveling at -100 km/h south. When and where do they meet. Remember to make
it directional and the one with the negative velocity must have a negative slope, therefore
starting at the 90 km mark.
Draw a position-time graph, using the slopes of the lines, you can roughly determine
when the two will pass.
4. Runner A gives B a 20 m head start head start in a race. A runs at 5.0 m/s and B at 3.0
m/s. Where and when will A catch B.
Draw a position-time graph, using the slopes of the lines, you can roughly determine
when the two will pass.
5. Slow Sam begins hiking at 12:00 at 2.5 km/h. Quick Quentin leaves 1.5 h later hiking
at 4 km/h. When and where does he catch Sam?
Draw a position-time graph, using the slopes of the lines, you can roughly determine
when the two will pass.
6. Student X jogs down a hallway at 5 m/s. Leaving at the same instant from the
opposite end of the 40 m long hallway Student Y walks quickly toward X at –2 m/s.
When and where do they meet?
Draw a position-time graph, using the slopes of the lines, you can roughly determine
when the two will pass.
7. Consider the graph at the right. The object whose motion is represented by this graph
is ... (include all that are true):
a. moving in the positive direction.
b. moving with a constant velocity.
c. moving with a negative velocity.
d. slowing down.
e. changing directions.
f. speeding up.
g. moving with a positive acceleration.
h. moving with a constant acceleration.
8. Describe the motion depicted by the following velocity-time graphs. In your
descriptions, make reference to the direction of motion (+ or - direction), the velocity and
acceleration and any changes in speed (speeding up or slowing down) during the various
time intervals (e.g., intervals A, B, and C).
a.
b.
c.
9. Determine the acceleration of the object depicted in the velocity-time graph at
intervals 0-4 seconds and 4-8 seconds
Time
Velocity
(s)
(m/s)
0
1
2
3
4
5
6
7
8
2
2
2
2
2
4
6
8
10
10. The velocity-time graph for a two-stage rocket is shown below. Use the graph and
your understanding of slope calculations to determine the acceleration and displacement
of the rocket during the listed time intervals.
a. t = 0 - 1 second
b. t = 1 - 4 second
c. t = 4 - 12 second
11. When lines on a velocity-time graph intersect, does it mean that the two cars are
passing by each other?
HOMEWORK LESSON 2 PART 2 KEY
1. t = 20 seconds, 60 m
2. t = 1.5 hours , 120 km
3. t = 0.5 hours, 40 km from the South (Penticton)
4. t = 10 seconds, 50 m
5. t = 4 hours, 10 km
6. t = 5.7 s, 28.6 m from student X
7. No Answer Key
8. No Answer Key
9. Find the slope of the lines, 0 4 s = 0 m/s2, 4 8 s = 2 m/s2
10. a. The acceleration is +40 m/s/s. The acceleration is found from a slope calculation.
The line rises +40 m/s for every 1 second of run. Displacement = 40 m/s x 1.0 s = 40 m
b. The acceleration is +20 m/s/s. The acceleration is found from a slope calculation.
The line rises +60 m/s for 3 seconds of run. The rise/run ratio is +20 m/s/s.
Displacement = 40 x 3 + ½ x 60 x 3 = 120 +90 =210 m
c. The acceleration is -20 m/s/s. The acceleration is found from a slope calculation.
The line rises -160 m/s for 8 seconds of run. The rise/run ratio is -20 m/s/s.
Displacement = ½ x 5 x 100 + ½ x 3 x –60 = 250 – 90 = 160 m remember, when
an object has a positive and then negative velocity, it is moving in opposite
directions, therefore the distances are subtracted
11. No Answer Key