But I think there isn't any line from the origin that would be equal to the tangent at t=10s. At leaks it looks like that.haruspex said:The line from the origin to the curve at t=10s is the same as the tangent at t=10s? I don't think so.
Right, so t=10s is not the point we're after - that was just an example.Davidllerenav said:But I think there isn't any line from the origin that would be equal to the tangent at t=10s. At leaks it looks like that.
It seems that t=16 would do.haruspex said:Right, so t=10s is not the point we're after - that was just an example.
Find a point where the two lines to that point, as we have discussed, would be the same.
Bingo.Davidllerenav said:It seems that t=16 would do.
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.haruspex said:Bingo.
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.