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Instantaneous velocity

  1. Sep 1, 2015 #1
    You sit at NASA to control the Mars rover across the Martian surface 2.0 x 10^8 km away. The communication travels at the speed of light between Earth and Mars, and the rover's top speed is 2.0 m/min. How far ahead in the rover's field of view you have to watch out for a Martian cliff?

    My first attempt was converting km to m to start the problem.

    Can anyone guide me how to solve the problem? Thanks!
     
  2. jcsd
  3. Sep 1, 2015 #2

    billy_joule

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    That's a start.
    You have a distance in metres and a velocity in m/s (speed of light) what is the travel time of communication?
     
  4. Sep 1, 2015 #3
    Is speed of light 299 792 458 m/s? If so, I would take 2.0 x 10^11 m divide by 299 792 458 m/s.
     
  5. Sep 1, 2015 #4

    billy_joule

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    Correct.
    You have the communication delay time, now what?
    What are you trying to find?
     
  6. Sep 1, 2015 #5
    I would convert my communication time (approximately about 667.13 s) to minute. From there, I multiply my min value by 2.0 m/min to get distance of rover traveled.
     
  7. Sep 1, 2015 #6

    billy_joule

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    Correct.
     
  8. Sep 1, 2015 #7
    I have 22.24 m for distance of rover. What else do I need to complete the problem?
     
  9. Sep 1, 2015 #8

    billy_joule

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    Apply some common sense.
    What you see in the video feed is 22.24 m behind the actual position. If you see a cliff and want to stop you'll travel 22.24 m before the 'stop' signal is received by the rover.
    With that in mind:
     
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