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

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.

Answer QQ20 Answer

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

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

Answer QQ21 Answer

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

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.

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

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

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?

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

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.

Answer QQ22

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