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Kinematics Motion Graphs. 1. Displacement – Time graphs a = uniform positive velocity b = c = d = Displacement Time a b c d a = uniform positive velocity.

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Presentation on theme: "Kinematics Motion Graphs. 1. Displacement – Time graphs a = uniform positive velocity b = c = d = Displacement Time a b c d a = uniform positive velocity."— Presentation transcript:

1 Kinematics Motion Graphs

2 1. Displacement – Time graphs a = uniform positive velocity b = c = d = Displacement Time a b c d a = uniform positive velocity b = decreasing velocity c = at rest d = uniform negative velocity

3 The gradient of a displacement - time graph is equal to the velocity of the object at any instant. gradient = Δy = Δ displacement = Δs = velocity Δx Δ time Δt Instantaneous velocity is given by the gradient of the tangent to the line at a certain time, t. ΔyΔy ΔxΔx time displacement t v = m = Δy Δx

4 2. Velocity – Time graphs a = uniform positive acceleration b = c = d = Velocity Time a b c d a = uniform positive acceleration b = non uniform positive acceleration c = uniform velocity d = uniform negative acceleration (perhaps a deceleration)

5 Displacement Time a b c d Velocity Time a b c d

6 The gradient of a velocity - time graph is equal to the acceleration of the object at any instant. gradient = Δy = Δ velocity = Δv = acceleration Δx Δ time Δt Instantaneous acceleration is given by the gradient of the tangent to the line at a certain time, t. The area under any section of a v-t graph is equal to the total displacement over that time period. (So you could have a negative area, indicating negative displacement) area = average velocity x time = v x Δt = Δs

7 3. Acceleration – Time graphs We are mainly interested in uniform acceleration: a = uniform positive acceleration b = uniform negative acceleration Acceleration Time a b

8 Bouncing Balls

9 Q. Two masses follow the paths shown. By sketching a velocity time graph for each, determine which one will reach the end of the track first? A B v t v t A. They both travel the same total distance, therefore the area under the v-t graph must be the same. Thus ball B must reach the end first!

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