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Speed in A Plane Mirror |
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| Nov28-12, 11:37 AM | #1 |
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Speed in A Plane Mirror
Suppose a driver his reversing his car with a speed of 2m/s, the driver sees in his rear-view mirror (which is a plane mirror) the image of a truck parked behind his car. What will be the speed at which the image of the truck appears the driver?
According to me, since the distance between the car and truck is equal to their distance in the mirror (because its a plane mirror), the speed at which the image appears should also be 2m/s, but the answer behind the Olympiad book says something else and I can't really figure out any other way to think about this problem. Thanks |
| Nov28-12, 11:41 AM | #2 |
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Say you are a distance 2D from the mirror. How far away does the image appear?
What if you were a distance D from the mirror? |
| Nov28-12, 11:47 AM | #3 |
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In the first case 4D. In the second case 2D.
Ah, I see.... Just as the car approaches the truck, so does the image of the car in the mirror, therefore the apparent distance covered is 2 times the actual distance covered by the car. So, can we conclude that since 2 = d/t, 2d/t= 2(2)=4 ? |
| Nov28-12, 12:26 PM | #4 |
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Speed in A Plane Mirror
What frame of reference are you trying to use?
From the point of view of the driver, the image in his rear-view mirror will approach at a speed of 2 meters per second. I agree with you on this. From the point of view of a physicist standing by with a pencil, a piece of paper and an optics textbook the virtual image in the mirror will move at a speed of 4 meters per second relative to the ground. From the point of view of the driver, the image in his rear view mirror will be approaching at an apparent speed of 2 meters per second as above while the ground viewed through his windshield is receding at a speed of 2 meters per second. This adds up to a closing rate of 4 meters per second. [The distinction between a speed and a closing rate is a not important if we're sticking to classical physics] |
| Nov28-12, 12:58 PM | #5 |
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What did the book say that differed? |
| Nov28-12, 05:07 PM | #6 |
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I am not too happy with these conclusions (or I may be misunderstanding).
I see it this way. The image distance is equal to the object distance therefore the image speeds will be the same. To a ground based observer, the two will be approaching each other with velocities of +2 and -2 respectively. To the driver, the image will be approaching at 4. I'm assuming that the motion is normal to the plane of the mirror, for simplicity. |
| Nov28-12, 06:20 PM | #7 |
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A bystander (at the truck, say) will see the car and its mirror approaching at 2m/s, and see the image as catching the car up at 2m/s, for a total of 4m/s. The driver is not being approached by the mirror, so only sees the 2m/s motion of the image relative to the mirror. |
| Nov29-12, 06:16 AM | #8 |
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