Respuesta :
Answer:
(a) 2 feet.
(b) 2 feet.
Step-by-step explanation:
We have been given that the velocity function [tex]v(t)=\frac{1}{\sqrt{t}}[/tex] in feet per second, is given for a particle moving along a straight line.
(a) We are asked to find the displacement over the interval [tex]1\leq t\leq 4[/tex].
Since velocity is derivative of position function , so to find the displacement (position shift) from the velocity function, we need to integrate the velocity function.
[tex]\int\limits^b_a {v(t)} \, dt[/tex]
[tex]\int\limits^4_1 {\frac{1}{\sqrt{t}}} \, dt[/tex]
[tex]\int\limits^4_1 {\frac{1}{t^{\frac{1}{2}}} \, dt[/tex]
[tex]\int\limits^4_1 t^{-\frac{1}{2}} \, dt[/tex]
Using power rule, we will get:
[tex]\left[\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}}\right] ^4_1[/tex]
[tex]\left[\frac{t^{\frac{1}{2}}}{\frac{1}{2}}}\right] ^4_1[/tex]
[tex]\left[2t^{\frac{1}{2}}\right] ^4_1[/tex]
[tex]2(4)^{\frac{1}{2}}-2(1)^{\frac{1}{2}}=2(2)-2=4-2=2[/tex]
Therefore, the total displacement on the interval [tex]1\leq t\leq 4[/tex] would be 2 feet.
(b). For distance we need to integrate the absolute value of the velocity function.
[tex]\int\limits^b_a |{v(t)|} \, dt[/tex]
[tex]\int\limits^4_1 |{\frac{1}{\sqrt{t}}}| \, dt[/tex]
Since square root is not defined for negative numbers, so our integral would be [tex]\int\limits^4_1 {\frac{1}{\sqrt{t}}} \, dt[/tex].
We already figured out that the value of [tex]\int\limits^4_1 {\frac{1}{\sqrt{t}}} \, dt[/tex] is 2 feet, therefore, the total distance over the interval [tex]1\leq t\leq 4[/tex] would be 2 feet.
