1. ## total dist. traveled

An object moves along a line so that its velocity at time $t$ is $v(t) = 1/2 + sin2t$ feet per second. Find the displacement and the total distance traveled by the object for $0 <= t <= 3pi/2$

i was able to find the displacement which is 3.35619 ft. need help on finding the total distance traveled.

2. Originally Posted by viet
An object moves along a line so that its velocity at time $t$ is $v(t) = 1/2 + sin2t$ feet per second. Find the displacement and the total distance traveled by the object for $0 <= t <= 3pi/2$

i was able to find the displacement which is 3.35619 ft. need help on finding the total distance traveled.
You already know that, in order to find displacement we integrate the velocity function:
$\Delta x = \int_0^{\frac{3\pi}{2}} dt \left ( \frac{1}{2} + sin(2t) \right )$

We do something similar to get the distance. Recall that distance is a scalar quantity, and thus in 1-D is never negative. This leads us to the formula:
$\Delta x = \int_0^{\frac{3\pi}{2}} dt \left | \frac{1}{2} + sin(2t) \right |$

In order to do this integral, we need to find the intervals over which the integrand is negative. (See graph below.) So we need to know when
$\frac{1}{2} + sin(2t) = 0$

$sin(2t) = -\frac{1}{2}$

$2t = \frac{7 \pi}{6}, \frac{11\pi}{6}$

$t = \frac{7 \pi}{12}, \frac{11\pi}{12}$

And we see that the integrand is negative for the interval $\left [ \frac{7 \pi}{12}, \frac{11\pi}{12} \right ]$

So:
$\Delta x = \int_0^{\frac{7\pi}{12}} dt \left ( \frac{1}{2} + sin(2t) \right ) -
\int_{\frac{7 \pi}{12}}^{\frac{11\pi}{12}} dt \left ( \frac{1}{2} + sin(2t) \right )$
$
+ \int_{\frac{11 \pi}{12}}^{\frac{3\pi}{2}} dt \left ( \frac{1}{2} + sin(2t) \right )$

I get $\Delta x = \sqrt{3} + \frac{5\pi}{12} + 1 \approx 4.04105$

-Dan

3. Hello, viet!

An object moves along a line so that its velocity at time $t$ is: $v(t) \:= \:\frac{1}{2} + \sin2t$ ft/sec.

Find the displacement and the total distance traveled by the object for $0 \leq t \leq \frac{3\pi}{2}$

Integrate: . $v(t)\:=\:\frac{1}{2} + \sin2t$ . . . and we have: . $s(t)\;=\;\frac{1}{2}t - \frac{1}{2}\cos2t + C$

. . Assume that the initial position is $s(0) = 0$, then $C = 0.$

. . Hence, the position function is: . $s(t) \;= \;\frac{1}{2}t - \frac{1}{2}\cos2t$

At $t = \frac{3\pi}{2}:\;s\left(\frac{3\pi}{2}\right)\;=\; \frac{1}{2}\left(\frac{3\pi}{2}\right) - \frac{1}{2}\cos(3\pi) \:=\:\frac{3\pi}{4} + \frac{1}{2} \:\approx\:2.8562$

Therefore, the displacement is $\boxed{2.8562}$ units to the right.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The object could have stopped and reversed direction.
. . This happens when $v(t) = 0.$

Let $v(t) \:=\:\frac{1}{2} + \sin2t\;=\;0$

We have: . $\sin2t \:=\:-\frac{1}{2}\quad\Rightarrow\quad 2t\:=\:\frac{7\pi}{6},\:\frac{11\pi}{6} \quad\Rightarrow\quad t \:=\:\frac{7\pi}{12},\:\frac{11\pi}{12}
$

. . $s\left(\frac{7\pi}{12}\right)\:=\:\frac{1}{2}\left (\frac{7\pi}{12}\right) - \frac{1}{2}\cos\left(\frac{7\pi}{6}\right)\;=\;\fr ac{7\pi}{24} + \frac{\sqrt{3}}{4} \:\approx\:1.3493$

. . $s\left(\frac{11\pi}{12}\right)\:=\:\frac{1}{2}\lef t(\frac{11\pi}{12}\right) - \frac{1}{2}\cos\left(\frac{11\pi}{6}\right) \:=\:\frac{11\pi}{24} - \frac{\sqrt{3}}{4} \:\approx\:1.0069$

Now we know all about the object's journey.

$\begin{array}{cccc} s(0) & = & 0\\ s(\frac{7\pi}{12}) & = & 1.3493\\ s(\frac{11\pi}{12}) & = & 1.0069\\ s(\frac{3\pi}{2}) & = & 2.8562\end{array}
\begin{array}{ccc} \}\;1.3493\text{ to the right} \\ \}\;0.3424\text{ to the left } \\ \}\;1.8493\text{ to the right}\end{array}$

. . Total distance: . $\boxed{3.541\text{ units}}$

But check my work . . . please!

4. There's only one problem with this:
Originally Posted by Soroban
. . Assume that the initial position is $s(0) = 0$, then $C = 0.$
We may certainly assume s(0) = 0, but then
$s(0) = \frac{0}{2} - \frac{1}{2}cos(2 \cdot 0) + C = 0$

gives
$C = \frac{1}{2}$, not $C = 0$

So all your distance calculations are off by 0.5 ft, which brings you back into agreement with my answers. (Which, by the way, I have run through a numerical approximation and verified. I never dare to assume you got your answer wrong! )

-Dan