# Homework Help: Simple Velocity Problem

1. Sep 8, 2010

### Allura

[PLAIN]http://lc.cit.jmu.edu/cgi-bin/plot.png?file=safavifn_jmu_1283993422_6827202_plot.data [Broken]

1. A bicyclist has the position-versus-time graph as shown. What is the bicyclist's velocity (in m/s) at t = 5 seconds?

2. Since this is a simple velocity problem, I thought it would be v= 60 m/ 5s

I don't know what I'm doing wrong, could it be that I'm reading the graph wrong?

Last edited by a moderator: May 4, 2017
2. Sep 8, 2010

### rock.freak667

velocity is the rate of change of displacement with time or v=ds/dt. So you need to find the gradient of that section from t= 0 to t=5. At what point does it cut the vertical axis? What position does t=5 correspond to?

3. Sep 8, 2010

### Allura

So would it be (61-51)/5? Since t=0 is 51m and t=5 is 61m?

4. Sep 8, 2010

### rock.freak667

Yes that looks correct to me.

5. Sep 8, 2010

### Allura

Thanks.

How would I do something like t=37s?

6. Sep 9, 2010

### rock.freak667

Then you'd need to consider the line segment between t=30 and t=40.

So get the position at t=37, that will be one point. Then get the position at any time between 30 to 40 (so t=31 for example). Then just use the same gradient formula you used before.

7. Sep 9, 2010

### Allura

Yay, I got it.

Thank you!