T Thatstudentm9 Member Joined Mar 4, 2017 Messages 46 Gender Male HSC 2018 Oct 24, 2018 #1 I feel embarrassed posting this but I don’t know how to do this question
HoldingOn Active Member Joined Dec 18, 2016 Messages 318 Location The Cosmos Gender Male HSC 2018 Oct 24, 2018 #2 Thatstudentm9 said: I feel embarrassed posting this but I don’t know how to do this question Click to expand... Displacement is represented by the area under the velocity curve. So we just calculate the area of the triangle we see i.e. 1/2 * 4 * 8= 16. But the particle starts at 2 so we have 16+2=18. Thus D Last edited: Oct 24, 2018
Thatstudentm9 said: I feel embarrassed posting this but I don’t know how to do this question Click to expand... Displacement is represented by the area under the velocity curve. So we just calculate the area of the triangle we see i.e. 1/2 * 4 * 8= 16. But the particle starts at 2 so we have 16+2=18. Thus D
W weaknuclearforce Member Joined Feb 6, 2016 Messages 80 Gender Male HSC N/A Oct 24, 2018 #3 To see the "displacement under the curve" explicitly you can observe integrating wrt t integrating wrt t You can see from the graph that maximum displacement must occur when velocity changes sign (direction) at t=4, or...
To see the "displacement under the curve" explicitly you can observe integrating wrt t integrating wrt t You can see from the graph that maximum displacement must occur when velocity changes sign (direction) at t=4, or...