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Homework Statement
Homework Equations
1/do+1/di=1/f
The Attempt at a Solution
I tried finding the distance of image at 27m and then at 30.5m and taking the difference but that didn't work.
How would I perform the calculation with instantaneous speed?Does that mean you tried to find the average speed of the image? I think their talking about "initial speed" indicates that they want instantaneous speed.
I tried doing this but unfortunately this didn't give me the correct answer. Is there a more accurate way of doing this?Your method gives the average velocity in the first second. It is a good approximation, if you calculate the image distances with enough significant digits (five at least). Show your work.
How do you get instantaneous speed (or more precisely, velocity) from position?How would I perform the calculation with instantaneous speed?
derivative?How do you get instantaneous speed (or more precisely, velocity) from position?
Ok cool, I'm still a little confused how i would apply it to this problem however.Yup.
What did you get? Yes, taking the derivative of di would be more accurate, but not much different.I tried doing this but unfortunately this didn't give me the correct answer. Is there a more accurate way of doing this?
You know the relationship between ##d_0## and ##d_i##: $$\frac 1 {d_o} + \frac 1 {d_i} = \frac 1 f.$$ Take the derivative with respect to t, of both sides of this equation, and you'll have a relationship between ##\frac {dd_o}{dt}## and ##\frac {dd_i}{dt}##.I'm still a little confused how i would apply it to this problem
With Δt = 1 s I get about a 13% difference in the final answer, using the full precision of my calculator.taking the derivative of di would be more accurate, but not much different.
Yes, it is true.With Δt = 1 s I get about a 13% difference in the final answer.
I am getting .00011 if i do it using the approximation with average velocity. It marks this as incorrect.With Δt = 1 s I get about a 13% difference in the final answer, using the full precision of my calculator.
I agree, if Dan shows us his working, we can tell him if he at least calculated that approximation correctly.
Yes, that's what I got. Another way to get more accuracy would be to use a smaller time interval. In principle, if you make it small enough, the answer will get close enough to the exact answer to make your software happy. But you have to be very careful to avoid roundoff errors.I am getting .00011 if i do it using the approximation with average velocity.
Great i ended up doing this and got the right answer. Thanks!Yes, that's what I got. Another way to get more accuracy would be to use a smaller time interval. In principle, if you make it small enough, the answer will get close enough to the exact answer to make your software happy. But you have to be very careful to avoid roundoff errors.