Physical experiments have confirmed that the body dropped from a tall cliff covers a distance of `4.9t^2` metres in t seconds, i.e., distance `s` in metres covered by the body as a function of time `t` in seconds is given by `s = 4.9t^2.`
The adjoining Table 13.1 gives the distance travelled in metres at various intervals of time in seconds of a body dropped from a tall cliff.
The objective is to find the veloctiy of the body at time `t = 2` seconds from this data. One way to approach this problem is to find the average velocity for various intervals of time ending at `t = 2` seconds and hope that these throw some light on the velocity at `t = 2` seconds.
Average velocity between `t = t_1` and `t = t_2` equals distance travelled between `t = t_1` and `t = t_2` seconds divided by `(t_2 – t_1).` Hence the average velocity in the first two seconds
`= ( "Distance travelled between "t = t_2" and "t = t_1 ) /["Time interval " (t_2 – t_1)]`
`= [(19.6-0)m]/[ (2-0)s]= 9.8 m//s`
Similarly, the average velocity between `t = 1` and `t = 2` is
`= [(19.6-4.7)m]/[ (2-1)s]= 14.7 m//s`
Likewise we compute the average velocitiy between `t = t_1` and `t = 2` for various `t_1.` The following Table `13.2` gives the average velocity `(v), t = t_1` seconds and `t = 2` seconds.
From Table 13.2, we observe that the average velocity is gradually increasing.
Hoping that nothing really dramatic happens between `1.99` seconds and `2` seconds, we conclude that the average velocity at `t = 2` seconds is just above `19.551m//s.`
Compute the average velocities for various time intervals starting at `t = 2` seconds. As before the average velocity `v` between `t = 2` seconds and `t = t_2` seconds is
`= ( "Distance travelled between 2 seconds and seconds " ) /["Time interval " (t_2 –2 )]`
`= ( "Distance travelled in seconds - Distance travelled in 2 seconds " ) / (t_2 –2 )`
The following Table `13.3` gives the average velocity `v` in metres per second between `t = 2` seconds and `t_2` seconds.
In the first set of computations, what we have done is to find average velocities in increasing time intervals ending at `t = 2` and then hope that nothing dramatic happens just before `t = 2.`
In the second set of computations, we have found the average velocities decreasing in time intervals ending at `t = 2` and then hope that nothing dramatic happens just after `t = 2`.
Purely on the physical grounds, both these sequences of average velocities must approach a common limit.
We can safely conclude that the velocity of the body at `t = 2` is between `19.551m//s` and `19.649 m//s.`
Technically, we say that the instantaneous velocity at `t = 2` is between `19.551 m//s` and `19.649 m//s`.
`color{green} "As is well-known, velocity is the rate of change of distance."`
From the given data of distance covered at various time instants we have estimated the rate of change of the distance at a given instant of time. We say that the derivative of the distance function `s = 4.9t^2` at `t = 2` is between `19.551` and `19.649.`
In the limit as the sequence of time intervals h1, h2 approaches
zero, the sequence of average velocities approaches the same limit as does the sequence of ratios
`(C_1B_1)/(AC_1), (C_2B_2)/(AC_2),(C_3B_3)/(AC_3)`
Physical experiments have confirmed that the body dropped from a tall cliff covers a distance of `4.9t^2` metres in t seconds, i.e., distance `s` in metres covered by the body as a function of time `t` in seconds is given by `s = 4.9t^2.`
The adjoining Table 13.1 gives the distance travelled in metres at various intervals of time in seconds of a body dropped from a tall cliff.
The objective is to find the veloctiy of the body at time `t = 2` seconds from this data. One way to approach this problem is to find the average velocity for various intervals of time ending at `t = 2` seconds and hope that these throw some light on the velocity at `t = 2` seconds.
Average velocity between `t = t_1` and `t = t_2` equals distance travelled between `t = t_1` and `t = t_2` seconds divided by `(t_2 – t_1).` Hence the average velocity in the first two seconds
`= ( "Distance travelled between "t = t_2" and "t = t_1 ) /["Time interval " (t_2 – t_1)]`
`= [(19.6-0)m]/[ (2-0)s]= 9.8 m//s`
Similarly, the average velocity between `t = 1` and `t = 2` is
`= [(19.6-4.7)m]/[ (2-1)s]= 14.7 m//s`
Likewise we compute the average velocitiy between `t = t_1` and `t = 2` for various `t_1.` The following Table `13.2` gives the average velocity `(v), t = t_1` seconds and `t = 2` seconds.
From Table 13.2, we observe that the average velocity is gradually increasing.
Hoping that nothing really dramatic happens between `1.99` seconds and `2` seconds, we conclude that the average velocity at `t = 2` seconds is just above `19.551m//s.`
Compute the average velocities for various time intervals starting at `t = 2` seconds. As before the average velocity `v` between `t = 2` seconds and `t = t_2` seconds is
`= ( "Distance travelled between 2 seconds and seconds " ) /["Time interval " (t_2 –2 )]`
`= ( "Distance travelled in seconds - Distance travelled in 2 seconds " ) / (t_2 –2 )`
The following Table `13.3` gives the average velocity `v` in metres per second between `t = 2` seconds and `t_2` seconds.
In the first set of computations, what we have done is to find average velocities in increasing time intervals ending at `t = 2` and then hope that nothing dramatic happens just before `t = 2.`
In the second set of computations, we have found the average velocities decreasing in time intervals ending at `t = 2` and then hope that nothing dramatic happens just after `t = 2`.
Purely on the physical grounds, both these sequences of average velocities must approach a common limit.
We can safely conclude that the velocity of the body at `t = 2` is between `19.551m//s` and `19.649 m//s.`
Technically, we say that the instantaneous velocity at `t = 2` is between `19.551 m//s` and `19.649 m//s`.
`color{green} "As is well-known, velocity is the rate of change of distance."`
From the given data of distance covered at various time instants we have estimated the rate of change of the distance at a given instant of time. We say that the derivative of the distance function `s = 4.9t^2` at `t = 2` is between `19.551` and `19.649.`
In the limit as the sequence of time intervals h1, h2 approaches
zero, the sequence of average velocities approaches the same limit as does the sequence of ratios
`(C_1B_1)/(AC_1), (C_2B_2)/(AC_2),(C_3B_3)/(AC_3)`