Skip to main content
added 23 characters in body; edited tags
Source Link
Qmechanic
  • 206.6k
  • 48
  • 566
  • 2.3k

Lets give an example. Velocity, v=ds/dt$v=ds/dt$. If we know the value of s $s$ (displacement) and t$t$ (time), we can instantly find the value of v$v$. But then this v$v$ will be the average velocity. Now consider, s= ut+.5at^2,$$s= ut+.5at^2,$$ which is equal to 100 meter. If a body travels 100 meter distance in 10 seconds, then we can say, v=s/t= (ut+.5at^2)/t = u+.5at.And$$v=s/t= (ut+.5at^2)/t = u+.5at.$$ And we know for sure this is average velocity. But when we consider ds/dt$ds/dt$, the formula becomes u+at$u+at$.

Here, u+at and u+.5at can never be equal, meaning average velocity in this case is not equal to actual velocity. But we know the answer for the given circumstance will be 10 meter per second velocity. So how this is incorrect? What did I miss? Can someone explain intuitively or logically and in details without mentioning formal definitions and jargon terms?

Lets give an example. Velocity, v=ds/dt. If we know the value of s (displacement) and t (time), we can instantly find the value of v. But then this v will be the average velocity. Now consider, s= ut+.5at^2, which is equal to 100 meter. If a body travels 100 meter distance in 10 seconds, then we can say, v=s/t= (ut+.5at^2)/t = u+.5at.And we know for sure this is average velocity. But when we consider ds/dt, the formula becomes u+at.

Here, u+at and u+.5at can never be equal, meaning average velocity in this case is not equal to actual velocity. But we know the answer for the given circumstance will be 10 meter per second velocity. So how this is incorrect? What did I miss? Can someone explain intuitively or logically and in details without mentioning formal definitions and jargon terms?

Lets give an example. Velocity, $v=ds/dt$. If we know the value of $s$ (displacement) and $t$ (time), we can instantly find the value of $v$. But then this $v$ will be the average velocity. Now consider, $$s= ut+.5at^2,$$ which is equal to 100 meter. If a body travels 100 meter distance in 10 seconds, then we can say, $$v=s/t= (ut+.5at^2)/t = u+.5at.$$ And we know for sure this is average velocity. But when we consider $ds/dt$, the formula becomes $u+at$.

Here, u+at and u+.5at can never be equal, meaning average velocity in this case is not equal to actual velocity. But we know the answer for the given circumstance will be 10 meter per second velocity. So how this is incorrect? What did I miss? Can someone explain intuitively or logically and in details without mentioning formal definitions and jargon terms?

Source Link

How does the result of derivative become different from average ratio calculation?

Lets give an example. Velocity, v=ds/dt. If we know the value of s (displacement) and t (time), we can instantly find the value of v. But then this v will be the average velocity. Now consider, s= ut+.5at^2, which is equal to 100 meter. If a body travels 100 meter distance in 10 seconds, then we can say, v=s/t= (ut+.5at^2)/t = u+.5at.And we know for sure this is average velocity. But when we consider ds/dt, the formula becomes u+at.

Here, u+at and u+.5at can never be equal, meaning average velocity in this case is not equal to actual velocity. But we know the answer for the given circumstance will be 10 meter per second velocity. So how this is incorrect? What did I miss? Can someone explain intuitively or logically and in details without mentioning formal definitions and jargon terms?