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Yes.Davidllerenav said:
Yes.Davidllerenav said:Ok! So for the second question, I'll have to calculate the maximum speed (slope) at the interval from t=10 sto t=14s, where the distance goes from s=1.0m to s-1.4m.
For the third, I must say that the instantaneous speed at t=16 is equal to the average speed from t=0s to t=16s.
That's all, right?
Thanks for your help!haruspex said:Yes.
You are welcome. You understood the method, yes?Davidllerenav said:Thanks for your help!
Yes. When I'm asked to find the maximum velocity on a graph of distance vs. time, I must look for the interval where the point travels more distance in the less time. When I'm asked to find the point where the instantaneous speed is equal to the average speed, I must find the point where its tangent line (slope/derivative) is equal to the line from the origin to that point, which is the average speed.haruspex said:You are welcome. You understood the method, yes?
Yes, but I meant more in terms of the principles behind that... that the tangent slope gives the instantaneous speed and the slope from the start point to current point gives current average speed.Davidllerenav said:When I'm asked to find the point where the instantaneous speed is equal to the average speed, I must find the point where its tangent line (slope/derivative) is equal to the line from the origin to that point, which is the average speed.
Yes. I understood that.haruspex said:Yes, but I meant more in terms of the principles behind that... that the tangent slope gives the instantaneous speed and the slope from the start point to current point gives current average speed.