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havj0y · Dec 12, 2011, 07:47 PM

a particle is moving along a straight line so that after t seconds after passing through a fixed point O on the line, its velocity is given by v(t)=t*sin((3^.5)*t)
a) find the values of t for which v(t)=0 given that 0 is less than or equal to t is less than or equal to 6

b) write down a mathematical expression for the total distance traveled by the particle in the particle in the first six seconds after passing through O. And find this distance.

Aurora2000 · Dec 13, 2011, 02:32 AM

$v(t):=t\sin(\sqrt{3}t)$

a) You want v(t)=0, then

$v(t):=t\sin(\sqrt{3}t)=0$

thus leading to t=0 or $\sqrt{3}t=n\pi \Longrightarrow t=\frac{n\pi}{\sqrt{3}}$ for some integer n: as you have imposed $0\leq t\leq 6$ , n can assume here only values 0, 1, 2, 3 (n=4 corresponds to $\sqrt{3}t=4\pi$ thus t>6).

Then you have $t=0, \frac{\pi}{\sqrt{3}}, \frac{2\pi}{\sqrt{3}}, \frac{3\pi}{\sqrt{3}}$

b) The distance (considering signs) traveled is

$s(T)=\int_0^T v(t)dt = \int_0^T t\sin (\sqrt{3}t) dt$

Integrating this expression you get

$\int_0^T t\sin (\sqrt{3}t) dt = -\frac{1}{\sqrt{3}}(-t\cos (\sqrt{3})+\sin (\sqrt{3}) )|_0^T$

and substituting $T=6$ you get the answer.

If you want the distance without considering signs then you have to integrate

$\int_0^T |v(t)|dt = \int_0^T |t\sin (\sqrt{3}t)| dt=$
$\int_0^{\frac{\pi}{\sqrt{3}}} v(t)dt - \int_{\frac{\pi}{\sqrt{3}}}^{\frac{2\pi}{\sqrt{3}} } v(t)dt <br /> +\int_{\frac{2\pi}{\sqrt{3}}}^{\frac{3\pi}{\sqrt{3 }}} v(t) dt-\int_{\frac{3\pi}{\sqrt{3}}}^{6} v(t)dt$

which is done in a similar manner, just remember that the primitive function of $t\sin(\sqrt{3}t)$ is $\frac{1}{\sqrt{3}}( -t\cos (\sqrt{3}t) +\sin(\sqrt{3}t) )$