Recent content by MihleDex

My bad, I used the formula for average velocity instead of the one for instantaneous velocity.

I'd say it's ##1/2aT^2## afar and the velocity is ##1/2aT## is there something I'm missing?

I combined the two graphs and drew the vertical line. I can see from the graph how the difference in displacement would be the difference in the area underneath the lines however I struggle to see the relationship to this expression: ##atT +1/2(a*T^2)##

Thanks a lot. The second term represents a distance and the velocity would be ## 1/2aT## So i think another way to represent it would be ## x = 1/2aT^2##

distance: atT +1/2(a*T^2) Sorry I verified this time using Dimensional analysis.

The difference would be 2tT + T^2

The 1st object has traveled for t - (-T) = t + T The 2nd object has traveled for t - 0 = t The 1st object has traveled as distance of: x = 1/2*a*(t+T)^2 The 2nd for: x = 1/2*a*(t)^2

Are your first question elaborating on what the homework question is asking?

From my understanding the the distance of the two at time t would be 0. Because at the statement it is said that both objects start at the origin. To me the two expressions mean the same if both objects start at the origin I don't see how it matters what value is assigned to t.