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How do you find the velocity of an object with respect to itself in a mirror?

How do you find the velocity of an object with respect to itself in a mirror?
 

Grade:12

1 Answers

Vikas TU
14149 Points
6 years ago
For any round mirror , the condition relating picture remove, question separation and span (or central length) is 
1u+1v=1f1u+1v=1f 
where the given terms have the typical significance concerning the current subject. 
Presently, let the given terms u and v be an element of time in their own particular separate means (i.e they can be distinctive capacity of times). 
Along these lines, u=f(t)u=f(t), v=h(t)v=h(t), f=constantf=constant (you can rather offer this to be a component of time yet that is a truly abnormal mirror you got there). 
Presently, take the first mirror condition and separate it as for time (we can on the grounds that we characterized it as a component of time), so 
−1u2∗ddt(u)+−vv2∗ddt(v)=0−1u2∗ddt(u)+−vv2∗ddt(v)=0 (since central length is a steady, unless you got an abnormal mirror). 
Improving the condition, we get 
−1u2∗ddt(u)=1v2∗ddt(v)−1u2∗ddt(u)=1v2∗ddt(v) 
Improving the squared terms to either side (I'll take them to the side of the picture speed) 
v2u2ddt(u)=ddt(v)v2u2ddt(u)=ddt(v) 
Presently, we have two terms which are separate elements of time. We for the most part aren't given both positions and henceforth assuming this is the case, at that point from the mirror condition, 
1v+1u=1f1v+1u=1f 
Increasing both sides by v, 
1+vu=vf⟹vu=f−vf1+vu=vf⟹vu=f−vf 
Substitution into the above speed condition gives you what you require (you can rather reworking the condition and afterward take for uvuv however that depends on your need). 
Additionally take note of that for plane mirrors, by taking f=∞f=∞, we get 
ddt(v)=−ddt(u)

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