Question

The velocity function, in feet per second, is given for a particle moving along a straight line. Find (a) the displacement and (b) the total distance that the particle travels over the given interval. v(t) = 1/√t, 1 ≤ t ≤ 4

Answer 1

Answer:

(a) 2 feet.

(b) 2 feet.

Step-by-step explanation:

We have been given that the velocity function $v(t)=\frac{1}{\sqrt{t}}$ 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 $1\leq t\leq 4$.

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.

$\int\limits^b_a {v(t)} \, dt$

$\int\limits^4_1 {\frac{1}{\sqrt{t}}} \, dt$

$\int\limits^4_1 {\frac{1}{t^{\frac{1}{2}}} \, dt$

$\int\limits^4_1 t^{-\frac{1}{2}} \, dt$

Using power rule, we will get:

$\left[\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}}\right] ^4_1$

$\left[\frac{t^{\frac{1}{2}}}{\frac{1}{2}}}\right] ^4_1$

$\left[2t^{\frac{1}{2}}\right] ^4_1$  

$2(4)^{\frac{1}{2}}-2(1)^{\frac{1}{2}}=2(2)-2=4-2=2$

Therefore, the total displacement on the interval  $1\leq t\leq 4$ would be 2 feet.

(b). For distance we need to integrate the absolute value of the velocity function.

$\int\limits^b_a |{v(t)|} \, dt$

$\int\limits^4_1 |{\frac{1}{\sqrt{t}}}| \, dt$

Since square root is not defined for negative numbers, so our integral would be $\int\limits^4_1 {\frac{1}{\sqrt{t}}} \, dt$.

We already figured out that the value of $\int\limits^4_1 {\frac{1}{\sqrt{t}}} \, dt$ is 2 feet, therefore, the total distance over the interval $1\leq t\leq 4$ would be 2 feet.