What is the speed of sprinter running away from lens?

[romeIAM]

Firsts of this is not my problem I don't even take physics but it is my girlfriends. She doesn't know how to get the answer an neither do I. I took some physics but not enough for this.

1. Homework Statement
A sports photographer has a 160-mm-focal-length lens on his camera. The photographer wants to photograph a sprinter running straight away from him at 5.5 m/s .

QUESTION:
What is the speed (in mm/s) of the sprinter's image at the instant the sprinter is 12 m in front of the lens?

Homework Equations

The Attempt at a Solution

I did some research online and through the book and I understand what it is asking. I visualize the problem in my head but I can't find the correct equations for this.

We just need some guidance on where to look or what equations to use and maybe how to use them.

Thank you physics forums community.

 

[vela]

Any relevant equations you can supply that might pertain to this problem?

 

[romeIAM]

Unfortunately at the moment I do not. Currently on khan academy after the Optics section maybe I will.
I apologize

 

[romeIAM]

Any relevant equations you can supply that might pertain to this problem?

The only equation i came out with was

1/f = (1/d object) + (1d image)

 

[vela]

Good. You're interested in the rate of change of $d_\text{obj}$ and $d_\text{img}$, so you need to differentiate that equation. This is like a related-rates problem in calculus.

 

[romeIAM]

Good. You're interested in the rate of change of $d_\text{obj}$ and $d_\text{img}$, so you need to differentiate that equation. This is like a related-rates problem in calculus.

so i got

1/f = 1/object + 1/ image = 1/160 = 1/12000-mm + 1/image =(37/6000)-mm is the image distance
so far am i correct?

 

[vela]

Not quite. You found $\frac{1}{d_\text{img}} = \frac{37}{6000\text{ mm}}$. You have to take the reciprocal of both sides to find $d_\text{img}$.

 

[romeIAM]

Not quite. You found $\frac{1}{d_\text{img}} = \frac{37}{6000\text{ mm}}$. You have to take the reciprocal of both sides to find $d_\text{img}$.

Yea i noticed it later. so ended up with . (6000/37) - mm which comes out to 162 mm.
so its 162mm/s

 

[vela]

No, you found the image distance. If the sprinter was stopped at a distance of 12.0 m, the image distance would remain at 162 mm. What you want to find is the rate of change of the image distance, which is going to be related to the speed of the sprinter. (You might have noticed you didn't use that piece of information yet.)