1. Velocity versus Time Instantaneous Velocity
Getting Velocity from the Position Graph
Outline

2. QQ20: Draw
Example
A rock is dropped straight down from a bridge and steadily speeds up as it falls. Draw the position versus time graph, as well as the associated velocity versus time graph. Think carefully about the signs!
Ignore air resistance.

3. Answer QQ20
Answer

4. Turning Points
Turning Points
When graphing, there are important points that can help you to quickly show the character of a curve
The point at which the position versus time graph goes from heading in one direction to heading in another is a turning point
A turning point on the position versus time graph is associated with a zero-crossing point on the velocity versus time graph
If you have a turning point at 5 seconds, you have a zero-crossing point at 5 seconds

5. QQ21: smooth curve
Example
Draw the velocity versus time graph that would be associated with the above position versus time graph.

6. Answer QQ21
Answer

7. Constant Acceleration
Going from Velocity to Position
Acceleration versus Time
Getting Acceleration from the Velocity Graph
Outline

8. Position from Velocity
Finding Position from Velocity Using a Graph
We can find the position of an object if we are given its starting position, as well as information about its velocity:
That means that if we know an objects initial position, we can use its velocity versus time graph to find its position at later times.

9. If Δx = vxΔt, how would we draw Δx on a graph of v vs. t?
v [m/s]
1m/s vx Δt
Δx = vxΔt
=(1 m/s)*10s
=10m
t [s]
10s

10. If Δx = vxΔt, how would we draw Δx on a graph of v vs. t?
v [m/s]
1m/s vx Δt
Δx = ½vxΔt
= ½(1m/s)(10s)
= 5m
t [s]
10s

11. Example: Finding position from a velocity versus time graph
Ex. Finding Velocity
Example: Finding position from a velocity versus time graph
If an object starts at an initial position x = 10 m, where is the object at t = 10 s?

12. Interval 1: Δx1 = (10m/s)(2s) = 20m
Ex. Finding Velocity 1 2 3 4 6 5
Interval 1: Δx1 = (10m/s)(2s) = 20m
Interval 2: Δx2 = 0.5(10m/s)(1s) = 5
Interval 3: Δx3 = (0m/s)(2s) = 0m
Interval 4: Δx4 = 0.5(-10m/s)(1s) = -5m
Interval 5: Δx5 = (-10m/s)(2s) = 20m
Interval 6: Δx6 = 0.5(-10m/s)(1s) = -5m
Overall: Δx = x0 + Δx1 + Δx2 + Δx3 + Δx4 + Δx5 + Δx6
= 10m + 20m + 5m + 0m – 5m + 20m – 5m
= 5m

13. QQ22: Find Velocity
Example
The object starts at an initial position x = 10 m. Draw its position versus time graph for the above time interval.

14. Answer QQ22

15. Do for next class: Read: sections 2.5, 2.6
Suggested problems: 2.13, 2.19
(no calculator: use g=10m/s2)