Posted by **Bean** on Monday, May 7, 2007 at 5:46pm.

Please could someone remind me how to calculate mean speed? I have to work out the mean speed of lava flow that travelled 13.5km over 21 days. I don't expect the answer, but some help on how to obtain it would be heplful! Thank you.

All you can do is average speed:

Average speed= distance/time

I suspect your teacher is using mean as average. Mean has a different meaning than this, but is often interchanged with average by laymen.

The mean speed would have nothing to do with distance and time, it would be the average of some speeds. For instance, if the speeds of the flow were measured as 11, 13, and 20 km/hr, the mean would be (11+13+20)/3 km/hr. That is a horse of a different color compared to average speed.

i need help on home work and bad its due tomorow and nothing is finished i really need help

Hi bean.Here is the solution.

13500m/(21*24*60*60)s =0.0074m/s.2Sig figs!

Or 0.64km*1000m =640m 2Sig figs!

One day is (24*60*60)s =86400s

So average speed =distance7time.

=640m/86400s =0.0074m/s.Good Luck

(I read off 13.5km also)

Hi bean.Here is the solution.

13500m/(21*24*60*60)s =0.0074m/s.2Sig figs!

Or 0.64km*1000m =640m 2Sig figs!

One day is (24*60*60)s =86400s

So average speed =distance7time.

=640m/86400s =0.0074m/s.Good Luck

(I read off 13.5km also)

SEE BEAN.OK

Do you not take the value given in the graph for the lava flow ... for EACH of the 21 days ????

i.e. 21 different values.. and divide that total by 21 ????

anyone ???

Do you not take the value given in the graph for the lava flow ... for EACH of the 21 days ????

i.e. 21 different values.. and divide that total by 21 ????

anyone ???

Do you not take the value given in the graph for the lava flow ... for EACH of the 21 days ????

i.e. 21 different values.. and divide that total by 21 ????

anyone ???

Good question Helper! I think you are right, BUT calculating the speed of the lava for each of the 21 days then calculating the mean speed feels like too much work for 5 marks. So, despite their use of the proper definition of mean in an earlier question, I calculated the average. I figure it should be worth 3 marks at least ;-)

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