1. 1D Motion graphs We will only consider straight line motion (also called ‘one dimensional’ or ‘linear’ motion). For straight line motion, direction can be indicated with a ‘+’ or ‘-’ sign as long as the positive direction has been specified. This PowerPoint show is designed to accompany (or remind you of) an oral presentation. Without the oral presentation some parts of this PowerPoint show may be meaningless.
2. Position-time graphs On a position-time graph, the gradient gives the velocity. This line has zero gradient, so the velocity is zero. Position / m Time / s 0 2 4 6 8 10 12 14 1 2 3 4 5
3. Line A has a gradient of 4/8 = 0.5 m/s. Line A is a steeper line -> higher velocity (forwards). On a position-time graph, the gradient gives the velocity. Line B has a gradient of 4/10 = 0.4 m/s. Position / m Time / s 0 2 4 6 8 10 12 14 1 2 3 4 5
4. Negative slope so negative velocity (going backwards). Velocity = gradient = -4/10 = -0.4 m/s Position / m Time / s 0 2 4 6 8 10 12 14 1 2 3 4 5
5. After the first 5 seconds the object moves back behind the zero position. The velocity is constant and negative: v = -12 / 10 = - 1.2 m/s Position / m Time / s 0 2 4 6 8 10 12 14 2 4 6 8 10 16 -2 -4 -6
6. What is the average velocity over the 8 seconds? Over the first six seconds? Over the period t = 8s to t=10s ? V = 4/8 = 0.5 m/s V = 2 / 6 = 0.333 m/s V = 2 / 2 = 1 m/s Position / m Time / s 0 2 4 6 8 10 12 14 1 2 3 4 5
7. What is the average velocity over the ten seconds? Over the first four seconds? Over the period t = 4s to t=10s ? V = 5/10 = 0.5 m/s V = 4/4 = 1 m/s V = 1/6 = 0.167 m/s Position / m Time / s 0 2 4 6 8 10 12 14 1 2 3 4 5
8. On a position-time graph, instantaneous velocity at any time is the slope of a tangent to the curve at that time. What is the instantaneous velocity at t= 2s? Instantaneous velocity = slope of tangent = (4.3-1.7) / 4 = 0.65 m/s X X Position / m Time / s 0 2 4 6 8 10 12 14 1 2 3 4 5
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10. Velocity-time graphs On a velocity-time graph, the gradient gives the acceleration . Here the gradient is zero, so the acceleration is zero. Velocity (m/s) Time / s 0 2 4 6 8 10 12 14 1 2 3 4 5
11. Acceleration = 4 / 10 = 0.4 m/s² On a velocity-time graph, the gradient gives the acceleration . Velocity (m/s) Time / s 0 2 4 6 8 10 12 14 1 2 3 4 5
12. In the first four seconds: acceleration = zero In the next 10 seconds: acceleration = -4 / 10 = -0.4 m/s² Velocity (m/s) Time / s 0 2 4 6 8 10 12 14 1 2 3 4 5
13. Velocity is negative (backwards). Acceleration = zero Velocity (m/s) Time / s 0 2 4 6 8 10 12 14 2 4 6 8 10 16 -2 -4 -6
14. In the first four seconds: acceleration = -2 / 4 = -0.5 m/s² In the next 6 seconds: acceleration = zero Velocity (m/s) Time / s 0 2 4 6 8 10 12 14 1 2 3 4 5 16 -1 -2 -3
15. Acceleration = -8 / 8 = -1 m/s² Note that the acceleration is constant here, whether the velocity is forwards, zero or backwards (negative). Velocity (m/s) Time / s 0 2 4 6 8 10 12 14 2 4 6 8 10 16 -2 -4 -6
16. In the first four seconds, a = 4 / 4 = 1 m/s² Between t= 4s and t = 10s, a = zero Between t= 10s and t = 12s, a = -4 / 2 = -2 m/s² Over the whole 12s, average acceleration = ZERO Velocity (m/s) Time / s 0 2 4 6 8 10 12 14 1 2 3 4 5
17. On a velocity-time graph, the area under the graph gives the displacement (the change in position). Displacement = 4 x 10 = 40 m Velocity (m/s) Time / s 0 2 4 6 8 10 12 14 1 2 3 4 5
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19. Δ x = ½ x 4 x 10 = 20m Velocity (m/s) Time / s 0 2 4 6 8 10 12 14 1 2 3 4 5
20. First four seconds: Δ x = 4 x 4 = 16m Next ten seconds: Δ x = ½ x 4 x 10 = 20m Total displacement = 16m + 20m = 36m Velocity (m/s) Time / s 0 2 4 6 8 10 12 14 1 2 3 4 5
21. Velocity (m/s) Time / s 0 2 4 6 8 10 12 14 2 4 6 8 10 16 -2 -4 -6 An area below the time axis is a NEGATIVE AREA and corresponds to a NEGATIVE displacement i.e. a backwards displacement . Δx = -4 x 8 = -32m
22. First four seconds: Δ x = ½ x -2 x 4 = -4m Next six seconds: Δ x = -2 x 6 = -12m Total displacement = -4m + -12m = -16m Velocity (m/s) Time / s 0 2 4 6 8 10 12 14 1 2 3 4 5 16 -1 -2 -3
23. Total displacement = 8m + -8m = 0m First four seconds: Δ x = ½ x 4 x 4 = 8m Next four seconds: Δ x = ½ x -4 x 4 = -8m THINK about that result! Velocity (m/s) Time / s 0 2 4 6 8 10 12 14 2 4 6 8 10 16 -2 -4 -6
24. Δx = 0.5 x 4 x 4 = 8m Δx = 0.5 x 4 x 2 = 4m Δx = 4 x 6= 24m Total displacement = 8m + 24m + 4m = 36m Velocity (m/s) Time / s 0 2 4 6 8 10 12 14 1 2 3 4 5
25. Equations of motion (1) Displacement = area on graph = ut Total displacement Δ x = ut + ½ (v-u)t = ut + ½vt - ½ut = ½(v+u)t = average velocity x time Velocity Time 0 t u v Displacement = area on graph = ½(v-u)t v-u
