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In the previous question, if the second car is overtaking at a relative speed of 15 ms^(-1), how fast will the image be moving ? |
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Answer» `1.2ms^(-1)` `(1)/(u) + (1)/(v) = (1)/(f) : (-1)/(u^(2))(du)/(dt) - (1)/(v^(2)) (dv)/(dt) = 0` `THEREFORE (dv)/(dt)=(v^(2))/(u^(2))(du)/(dt)=((19.4)/(600))^(2)xx15` |
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