Respuesta :
Answer:
The total distance that the turtle traveled during the 5 seconds recorded is [tex]Distance \:traveled\approx 3.838\:\frac{m}{s}[/tex]
Step-by-step explanation:
To estimate distance traveled of an object moving in a straight line over a period of time, from discrete data on the velocity of the object, we use a Riemann Sum. If we have a table of values
[tex]\left\begin{array}{ccccccc}time\:=\:t_i&t_0=0&t_1&t_2&...&t_n\\velocity\:=\:v(t_i)&v(t_0)&v(t_1)&v(t_2)&...&v(t_n)\end{array}\right[/tex]
where [tex]\Delta t=t_i-t_{i-1}[/tex], then we can approximate the displacement on the interval [tex][t_{i-1},t_i][/tex] by [tex]v(t_{i}) \times\Delta t[/tex].
Therefore the distance traveled of the object over the time interval [tex][0,t_n][/tex] can be approximated by
[tex]Distance \:traveled\approx |v(t_1)|\Delta t+|v(t_2)|\Delta t+...+|v(t_n)|\Delta t[/tex]
This is the right endpoint approximation.
We are given a table of values for v(t)
[tex]\left\begin{array}{cccccccc}t(sec)&0&1&2&3&4&5\\v(t)&0.078&0.83&0.75&0.98&0.853&0.425\end{array}\right[/tex]
Applying the right endpoint approximation formula we get,
[tex]\Delta t = 1\sec[/tex]
[tex]Distance \:traveled\approx |v(t_1)|\Delta t+|v(t_2)|\Delta t+|v(t_3)|\Delta t+|v(t_4)|\Delta t+|v(t_5)|\Delta t\\\\Distance \:traveled\approx 0.83(1)+0.75(1)+0.98(1)+0.853(1)+0.425(1)\\\\Distance \:traveled\approx 3.838\:\frac{m}{s}[/tex]
The total distance that the turtle traveled during the 5 seconds recorded is [tex]Distance \:traveled\approx 3.838\:\frac{m}{s}[/tex]
