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.
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.