1. Days 5 - 6
UNIT 1 Motion Graphs x t
Lyzinski Physics

2. Day #5. Acceleration on x-t graphs. v-t graphs
Day #5 * Acceleration on x-t graphs * v-t graphs * using v-t graphs to get a

3. x-t ‘s x t UNIFORM Velocity Speed increases as slope increases x t
Object at REST x t
Moving forward or backward x t
x-t ‘s
Object Positively Accelerating x t x t
Changing Direction x t
Object Speeding up
Object Negatively Accelerating

4. POSITIVE OR NEGATIVE ACCEL? SPEEDING UP OR SLOWING DOWN?x t x t
Slope of the tangent gives vinst
Getting more sloped  speeding up
Getting more + sloped  + Accel
Getting less sloped  slowing down
Slopes are getting less +  - Accel x t x t
Getting less sloped  slowing down
Slopes are getting less –  + Accel
Getting more sloped  speeding up
Slopes are getting more –  - Accel

5. An easy way to remember it
I’m Positive!!!
I’m Negative!!!

6. Find the acceleration in each case.
v1 = 10 m/s, v2 = 20 m/s, Dt = 5sec
v1 = 10 m/s, v2 = -20 m/s, Dt = 10sec
v1 = -9 km/h, v2 = -27 km/h, Dt = 3 h
v1 = -9 km/h, v2 = 6 km/h, Dt = 3 h

7. v-t ‘s UNIFORM Positive (+) Acceleration
Acceleration increases as slope increases v t v t
Changing Direction
v-t ‘s
UNIFORM Velocity
(no acceleration) v t v t
Object at REST
UNIFORM Negative (-) Acceleration

8. v-t graphs v(m/s) Constant + accel (slowing down) At rest t (s)
t (s) 8 6 4 2 -2 -4
At rest
Constant + accel (slowing down)
Constant - Vel
Constant negative accel (speeding up)
Constant negative accel (slowing down)
Constant + Vel (constant speed)
Constant + accel (speeding up)

9. How to get the velocity (v) at a certain time off a v-t graph
v(m/s)
t (s) 8 6 4 2 -2 -4
Example:
What is the velocity at t = 8 seconds?
Go over to t = 8.
Find the pt on the graph.
-2 m/s
Find the v value for this time.

10. Finding the average acceleration on a v-t graph
v(m/s) 8 6 4 2 -2 -4
Example:
What is the average acceleration between 0 & 2, 2 & 4, and 4 & 10 seconds?
a0-2 = (v2 – v1) / Dt
= rise / run
= +4/2 = +2 m/s2
A2-4 = (v2 – v1) / Dt
= rise / run
= 0 m/s2
A4-10 = (v2 – v1) / Dt
= rise / run
= -7 / 6 = m/s2

11. v-t graphs Slope of any segment is the AVERAGE acceleration v (m/s)
The slope of the tangent to the curve at any point is the INSTANTANEOUS acceleration
t (sec)
t t1

12. Open to in your Unit 1 packet 330 20 10
-10
-20
-30
v = -30 m/s s = 30 m/s
a = slope = (-30 m/s) / 16sec = m/s2
a = slope = (+57 m/s) / 32sec = m/s2 (approx)

13. Open to in your Unit 1 packet 330 20 10
-10
-20
-30 3) 4) 5)
You can’t say. You know its speed at the start, but not where it is 
Object is at rest whenever it crosses the t-axis  t = 0, 36, 80 sec
Const – accel (object speeds up), const – vel, const + accel (slows down), const + accel (speeds up), const – accel (slows down)

14. Day #6. v-t graphs. slopes & areas of v-t graphs
Day #6 * v-t graphs * slopes & areas of v-t graphs * instantaneous accelerations

15. x-t ‘s x t UNIFORM Velocity Speed increases as slope increases x t
Object at REST x t
Moving forward or backward x t
x-t ‘s
Object Positively Accelerating x t x t
Changing Direction x t
Object Speeding up
Object Negatively Accelerating

16. v-t ‘s UNIFORM Positive (+) Acceleration
Acceleration increases as slope increases v t v t
Changing Direction
v-t ‘s
UNIFORM Velocity
(no acceleration) v t v t
Object at REST
UNIFORM Negative (-) Acceleration

17. A Quick Review
The slope between 2 points on an x-t graph gets you the _______________.
The slope at a single point (the slope of the tangent to the curve) on an x-t graph gets you the ____________.
The slope between 2 points on a v-t graph gets you the ____________.
The slope at a single point (the slope of the tangent to the curve) on a v-t graph gets you the ____________.
Average velocity
Inst. velocity
Avg. accel.
Inst. accel.

18. NEW CONCEPT
When you find the area “under the curve” on a v-t graph, this gets you the displacement during the given time interval. v t v t
The “area under the curve” is really the area between the graph and the t-axis.
This is NOT the area under the curve 

19. Find the area under the curve from …. 0-4 seconds. 4-6 6-10 0-10
The displacement during the first 4 seconds is -20m
A = ½ (4)(-10) = -20m
A = ½ (2)(-10) = -10m
The displacement during the next 2 seconds is -10m
A = ½ (4)(15) = 30m
The displacement during the next 4 seconds is 30m v t 15
-10
A = (-10) + 30
= 0m
The OVERALL displacement from 0 to 10 seconds is zero (its back to where it started)

20. How to find the displacement from one time to another from a v-t graph
v(m/s)
t (s) 8 6 4 2 -2 -4
Example:
What is the displacement from t = 2 to t = 10?
Find the positive area bounded by the “curve”
12 m + 4 m = 16 m
Find the negative area bounded by the “curve”
(-2.25 m) + (-4.5 m) = m
Add the positive and negative areas together 
Dx = 16 m + (-6.75 m) = 9.25 m

21. How to find the distance traveled from one time to another from a v-t graph
v(m/s)
t (s) 8 6 4 2 -2 -4
Example:
What is the distance traveled from t = 2 to t = 10?
Find the positive area bounded by the “curve”
12 m + 4 m = 16 m
Find the negative area bounded by the “curve”
(-2.25 m) + (-4.5 m) = m
Add the MAGNITUDES of these two areas together 
distance = 16 m m = m

22. How to find the average velocity during a time interval on a v-t graph
v(m/s)
t (s) 8 6 4 2 -2 -4
Example:
What is the average velocity from t = 2 to t = 10?
The DISPLACEMENT is simply the area “under” the curve.
Dx = 16 m + (-6.75 m)
= 9.25 m
12 m + 4 m = 16 m
(-2.25 m) + (-4.5 m) =
The AVG velocity = Dx / Dt = 9.25 m / 8 s = 1.22 m/s

23. How to find the final position of an object using a v-t graph (and being given the initial position)
v(m/s)
t (s) 8 6 4 2 -2 -4
Example:
What is the final position after t = 10 seconds if xi = 40 m?
The DISPLACEMENT during the 1st 10 sec is simply the area “under” the curve.
Dx = 20 m + (-6.75 m)
= m
4 m 12 m + 4 m = 20 m
(-2.25 m) + (-4.5 m) =
Dx = x2 – x1  x2 = Dx + x1 = m + 40 m = m

24. v-t graphs Slope of any segment is the AVERAGE acceleration v (m/s)
The slope of the tangent to the curve at any point is the INSTANTANEOUS acceleration
The area under the curve between any two times is the CHANGE in position (the displacement) during that time period.
t (sec)
t t1

25. Open to in your Unit 1 packet 330 20 10
-10
-20
-30 6) 7)
Distance travelled = |area| = | ½ (16)(-30) | = 240m
s = d/t = 240 m / 16 sec = 15 m/s
Displacement = |area| = ½ (16)(-30) + 8 (-30) = -480m
v = Dx/t = -480 m / 24 sec = -20 m/s

26. Open to in your Unit 1 packet 330 20 10
-10
-20
-30 8)
Find all the areas “under the curve” from 0 to 44 sec
Area = ½ (16)(-30) + 12(-30) + ½ (8)(-30) + ½ (8)(30) = m
Area = Dx = m  Dx = x2 – x1  m = x2 – (-16m)  x2 = - 616m

27. Open to in your Unit 1 packet 4
+3.3 m/s2
+10 m/s
0 m/s
+75 m

28. 10
-10 5
-2 m/s2
0 m
50 m
30 m

29. 14 & 34 sec +.35 m/s2 + 8 m/s + 2 m/s2 approx 0.8 m/s2 9) 10) 11) 12)
13)
14 & 34 sec
+.35 m/s2
+ 8 m/s
+ 2 m/s2
approx 0.8 m/s2