Download presentation

Presentation is loading. Please wait.

Published bySofia Atkinson Modified over 4 years ago

1
Graphical Analysis of Linear Motion

2
A car travels along a road at a constant velocity of 10. m/s time (s) position (m) 0 0 1 10 2 20 3 30

3
slope = Δx/Δt ΔtΔt ΔxΔx = 20 m/2 s = 10 m/s velocity

4
position vs. time graph to find displacement: to find velocity: subtract values from graph find slope if straight line acceleration = 0

5
Displacement for 1 st 3 seconds: 30 m area under graph slope:0acceleration

6
v vs. t graph to find displacement: to find velocity: to find acceleration: find area read graph find slope

7
Object dropped from a tall building (use g = 10 m/s 2 ) time (s) position (m) 0 0 1 5 2 20 3 45 x = x 0 + v 0 t + ½ at 2 x 0 = 0; v 0 = 0

8
velocity @ 2 s: slope of tangent line slope = 20/1 = 20 m/s

9
time (s) velocity (m/s) 0 0 1 10 2 20 3 30 v = v 0 + at

10
acceleration @ 2 s:slope 10 m/s 2 displacement for 1 st 3 s:area Δx = 45 m

11
area:30 m/s= Δv

12
Graphs (vs. time) position velocityacceleration slope area

13
If acceleration is positive (constant); x 0 = 0; v 0 = 0 x = ½ at 2

14
To produce a straight-line graph: slope = ½ a also: t 2 vs. x, x vs. t, t vs. x

20
Graphs of x, v, a

Similar presentations

OK

CH. 2 NOTES Abbreviated. Distance vs. Displacement Distance – how far an object travels (meters) Displacement – straight line path between two points.

CH. 2 NOTES Abbreviated. Distance vs. Displacement Distance – how far an object travels (meters) Displacement – straight line path between two points.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google