Question

At the moment t=0 a particle leaves the origin and moves in the positive direction of the x axis its velocity varies with time as $=v_0(1-t/\tau ),where{v}_0$ is the initial velocity vector whose modulus equals $v_0=10.0cm/s;\tau =5.0s$. Find :
(a) the x coordinate of the principle at the moments of time 6.0, 10 and 20 s;
(b) the moments of time when the particle is at the distance 10.0 cm from the origin;

Answer 1

Given,

$velocity=V_0(1-\frac{t}{\tau })$

Where $v_0$ is the initial velocity vector.

$V_0=10cm/s$

So,

a)As the particle lives the origin

At $t=0$

So,

$\Delta x=x=\int v_{x}dt........1$

$\overrightarrow{V}=V_0(1-\frac{t}{\tau })$

Where ${\overrightarrow{V}}_0$ is the directed toward the positive axis

So,

$V_x=V_0(1-\frac{t}{\tau }).......2$

From equation 1 and 2 we get,

$x={\int }_0^t{V}_0(1-\frac{t}{\tau })dt$

$=V_{0}t(1-\frac{t}{2\tau })........3$

Thus the x-coordinate of the particle at $t=66s$

$x=10\times 6(1-\frac{6}{10})=24cm$

Similarly,

At $t=10s$

$x=10\times 10(1-\frac{10}{10})$

$=0$

And at $t=20s$

$x=10\times 20(1-\frac{20}{10})$

$-200cm$

b)

at the moments of the particle is at the distance of 10cm from the origin,

$x=\pm 10cm$

Put this value in equation # then we get,

$10=10t(1-\frac{t}{10})$

$t^2-10t+10=0$

$t=\frac{10\pm \sqrt{100+40}}{2}=5\pm \sqrt{15}s$

Now put $x=-10cm$ in equation 3 then we get,

$-10=10(1-\frac{t}{10})$

By solving we get,

$t=5\pm \sqrt{35}s$