Timeline for How does the result of derivative become different from average ratio calculation?
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| May 22 at 17:43 | comment | added | Luke Pritchett | There is more than one way to travel 100m in 10 seconds so of course there should be more than one "answer." For example, I could run really fast for 5 seconds then walk for 5 seconds. Or I could walk for 5 seconds and then run fast for 5 seconds. Or I could run 200 in 5 seconds and then run backwards for 5 seconds. There is no single number that can describe all these different options. That's why we have different ways of talking about velocity. The derivative gives me one piece of information about my motion, the average velocity gives a different piece of information. | |
| May 22 at 13:50 | comment | added | Arafat | My doubt is derivatives give one answer and average velocity v=s/t gives another. Why is so when I specifically say that a body travelled 100m distance in 10 sec? | |
| May 22 at 13:49 | comment | added | Arafat | Sorry for the confusion. What I meant is that v=s/t is the average velocity and v= ds/dt is the general definition. But these two do not match according to what I stated. I really need to understand this. Please ask question if you do not understand or confused about I tried to mean. | |
| May 22 at 12:54 | history | answered | Luke Pritchett | CC BY-SA 4.0 |