# Instantaneous velocity

1. Sep 1, 2015

### Manh

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. Sep 1, 2015

### billy_joule

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?

3. Sep 1, 2015

### Manh

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.

4. Sep 1, 2015

### billy_joule

Correct.
You have the communication delay time, now what?
What are you trying to find?

5. Sep 1, 2015

### Manh

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.

6. Sep 1, 2015

Correct.

7. Sep 1, 2015

### Manh

I have 22.24 m for distance of rover. What else do I need to complete the problem?

8. Sep 1, 2015

### billy_joule

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: