1. Motion in One Dimension (Velocity vs. Time) Chapter 5.2

2. What is instantaneous velocity?

3. What effect does an increase in velocity have on displacement?

4. Determining instantaneous velocity
1997 World Championships - Athens, Greece
Maurice Green
100 90 80
y = 11.65x R
= 1.00 2 70 60
Distance (m) 50 40 30 20
y = 1.13x
+ 4.08x R 2
= 1.00 10 2 4 6 8 10
Time (s)

5. How do you determine the instantaneous velocity?
What is the runners velocity at t = 1.5s?
Instantaneous velocity = slope of line tangent to curve.

6. Determining the instantaneous velocity from the slope of the curve.
m = rise/run
m = 25m – 5 m
3.75s – 1.0s
m = 7.3 m/s
v = s

7. Instead of position vs. time, consider velocity vs. time.
Relatively constant
velocity
High acceleration

8. How can displacement be determined from a v vs. t graph?
Measure the area under the curve.
d = v*t
Where
t is the x component
v is the y component
Time
Velocity A2 A1
A1 = d1 = ½ v1*t1
A2 = d2 = v2*t2
dtotal = d1 + d2

9. Measuring displacement from a velocity vs. time graph.
A = b x h
A = (7.37)(11.7)
A = 86.2 m
A = ½ b x h
A = ½ (2.36)(11.7)
A = 13.8 m

10. What information does the slope of the velocity vs. time curve provide?
Positively sloped curve = increasing velocity (Speeding up).
Negatively sloped curve = decreasing velocity (Slowing down).
Horizontally sloped curve = constant velocity.
Time
Velocity
Positive Acceleration A
Time
Velocity
Negative Acceleration B
Time
Velocity
Zero Acceleration C

11. What is the significance of the slope of the velocity vs. time curve?
Since velocity is on the y-axis and time is on the x-axis, it follows that the slope of the line would be:
Therefore, slope must equal acceleration.
Time
Velocity

12. Acceleration determined from the slope of the curve.
What is the acceleration at t = 2 seconds?
rise
run
vf – vi
tf – ti
13m/s-7m/s
3.75s-0.75s
m = 2.0 m/s2
Since m = a:
a = 2.0 m/s2
m =
Slope =

13. Determining velocity from acceleration
If acceleration is considered constant:
a = v/t = (vf – vi)/(tf – ti)
Since ti is normally set to 0, this term can be eliminated.
Rearranging terms to solve for vf results in:
vf = vi + at
Time
Velocity
Positive Acceleration
Velocity

14. Position, velocity and acceleration when t is unknown.
d = di + ½ (vf + vi)*t (1)
vf = vi + at (2)
Solve (2) for t: t = (vf – vi)/a and substitute back into (1)
df = di + ½ (vf + vi)(vf – vi)/a
By rearranging:
vf2 = vi2 + 2a*(df – di) (3)

15. Substitute (2) into (1) for vf df = di + ½ (vi + at + vi)*t
Alternatively, If time and acceleration are known, but the final velocity is not:
df = di + ½ (vf + vi)*t (1)
vf = vi + at (2)
Substitute (2) into (1) for vf
df = di + ½ (vi + at + vi)*t
df = di + vit + ½ at2 (4)

16. Formulas for Motion of Objects
Equations to use when an accelerating object has an initial velocity.
Form to use when accelerating object starts from rest (vi = 0).
d = ½ (vi + vf) t
d = ½ vf t
vf = vi + at
vf = at
d = vi t + ½ a(t)2
d = ½ a(t)2
vf2 = vi2 + 2ad
vf2 = 2ad

17. Acceleration due to Gravity
All falling bodies accelerate at the same rate when the effects of friction due to water, air, etc. can be ignored.
Acceleration due to gravity is caused by the influences of Earth’s gravity on objects.
The acceleration due to gravity is given the special symbol g.
The acceleration of gravity is a constant close to the surface of the earth.
g = 9.81 m/s2

18. Example 1: Calculating Distance
A stone is dropped from the top of a tall building. After 3.00 seconds of free-fall, what is the displacement, y of the stone?
Data y ?
a = g
-9.81 m/s2 vf
n/a vi
0 m/s t
3.00 s

19. Example 1: Calculating Distance
From your reference table:
d = vit + ½ at2
Since vi = 0 we will substitute g for a and y for d to get:
y = ½ gt2
y = ½ (-9.81 m/s2)(3.00 s)2
y = m

20. Example 2: Calculating Final Velocity
What will the final velocity of the stone be?
Data y
-44.1 m
a = g
-9.81 m/s2 vf ? vi
0 m/s t
3.00 s

21. Example 2: Calculating Final Velocity
Using your reference table:
vf = vi + at
Again, since vi = 0 and substituting g for a, we get:
vf = gt
vf = (-9.81 m/s2)(3.00 s)
vf = m/s
Or, we can also solve the problem with:
vf2 = vi2 + 2ad, where vi = 0
vf = [(2(-9.81 m/s2)(44.1 m)]1/2

22. Example 3: Determining the Maximum Height
How high will the coin go?
Data y ?
a = g
-9.81 m/s2 vf
0 m/s vi
5.00 m/s t

23. Example 3: Determining the Maximum Height
Since we know the initial and final velocity as well as the rate of acceleration we can use:
vf2 = vi2 + 2ad
Since Δd = Δy we can algebraically rearrange the terms to solve for Δy.

24. Example 4: Determining the Total Time in the Air
How long will the coin be in the air?
Data y
1.28 m
a = g
-9.81 m/s2 vf
0 m/s vi
5.00 m/s t ?

25. Example 4: Determining the Total Time in the Air
Since we know the initial and final velocity as well as the rate of acceleration we can use:
vf = vi + aΔt, where a = g
Solving for t gives us:
Since the coin travels both up and down, this value must be doubled to get a total time of 1.02s

26. Key Ideas
Instantaneous velocity is equal to the slope of a line tangent to a position vs. time graph.
Slope of a velocity vs. time graphs provides an objects acceleration.
The area under the curve of a velocity vs. time graph provides the objects displacement.
Acceleration due to gravity is the same for all objects when the effects of friction due to wind, water, etc can be ignored.

27. Important equations to know for uniform acceleration.
df = di + ½ (vi + vf)*t
df = di + vit + ½ at2
vf2 = vi2 + 2a*(df – di)
vf = vi +at
a = Δv/Δt = (vf – vi)/(tf – ti)

28. Displacement when acceleration is constant.
Displacement = area under the curve.
Δd = vit + ½ (vf – vi)*t
Simplifying:
Δd = ½ (vf + vi)*t
If the initial position, di, is not 0, then:
df = di + ½ (vf + vi)*t
By substituting vf = vi + at
df = di + ½ (vi + at + vi)*t
df = di + vit + ½ at2 vf
d = ½ (vf-vi)t
d = vit vi t