● 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 (t2 – t1). Hence the average velocity in the first two seconds
`= ("Distance travelled between" t_2 = 2 "and" t_1 = 0)/("Time interva"l (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 - 4.9 )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. As we make the time intervals ending at t = 2 smaller, we see that we get a better idea of the velocity at t = 2. 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 19.551 m/s.
●This conclusion is somewhat strengthened by the following set of computation. 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" )/(t_2 -2)`
`= ("Distance travelled in" t_2 "seconds - Distance travelled in 2 seconds")/(t_2 -2)`
`= ("Distance travelled in seconds - 19.6")/(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.
● Here again we note that if we take smaller time intervals starting at `t = 2,` we get better idea of the velocity at `t = 2.`
● 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 19.551 m/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. As is well-known, velocity is the rate of change of displacement. Hence what we have accomplished is the following.
● 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.
●An alternate way of viewing this limiting process is shown in Fig 13.1. This is a plot of distance s of the body from the top of the cliff versus the time t elapsed. In the limit as the sequence of time intervals `h_1, h_2`, ..., approaches zero, the sequence of average velocities approaches the same limit as does the sequence of ratios
`color{lime} {(C_1B_1)/(AC_1) , (C_2B_2)/(AC_2), (C_3B_3)/(AC_3), .......}`
● where `color{green} {C_1B_1 = s_1 – s_0}` is the distance travelled by the body in the time interval `h_1 = AC_1`, etc. From the Fig 13.1 it is safe to conclude that this latter sequence approaches the slope of the tangent to the curve at point A. In other words, the instantaneous velocity `v(t)` of a body at time t = 2 is equal to the slope of the tangent of the curve `s = 4.9t^2` at t = 2.
● 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 (t2 – t1). Hence the average velocity in the first two seconds
`= ("Distance travelled between" t_2 = 2 "and" t_1 = 0)/("Time interva"l (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 - 4.9 )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. As we make the time intervals ending at t = 2 smaller, we see that we get a better idea of the velocity at t = 2. 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 19.551 m/s.
●This conclusion is somewhat strengthened by the following set of computation. 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" )/(t_2 -2)`
`= ("Distance travelled in" t_2 "seconds - Distance travelled in 2 seconds")/(t_2 -2)`
`= ("Distance travelled in seconds - 19.6")/(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.
● Here again we note that if we take smaller time intervals starting at `t = 2,` we get better idea of the velocity at `t = 2.`
● 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 19.551 m/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. As is well-known, velocity is the rate of change of displacement. Hence what we have accomplished is the following.
● 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.
●An alternate way of viewing this limiting process is shown in Fig 13.1. This is a plot of distance s of the body from the top of the cliff versus the time t elapsed. In the limit as the sequence of time intervals `h_1, h_2`, ..., approaches zero, the sequence of average velocities approaches the same limit as does the sequence of ratios
`color{lime} {(C_1B_1)/(AC_1) , (C_2B_2)/(AC_2), (C_3B_3)/(AC_3), .......}`
● where `color{green} {C_1B_1 = s_1 – s_0}` is the distance travelled by the body in the time interval `h_1 = AC_1`, etc. From the Fig 13.1 it is safe to conclude that this latter sequence approaches the slope of the tangent to the curve at point A. In other words, the instantaneous velocity `v(t)` of a body at time t = 2 is equal to the slope of the tangent of the curve `s = 4.9t^2` at t = 2.