# A sphere is fired downwards into a medium

A sphere is fired downwards into a medium with an initial speed of 27 m/s. If it experiences a deceleration of $a = (-6t)$ m/$s^2$ where t is in seconds, determine the distance traveled before it stops.

#### Solution:

When acceleration is given with respect to time, we can write it as:

$a(t)=\dfrac{dv}{dt}$

solving for dv yields:

$dv=a(t)\,dt$

We can now take the integral of both sides to find the time when the sphere stops.

$\,\displaystyle \int^{v}_{v_0}dv=\int^{t}_{t_0}a(t)\,dt$

Substituting our acceleration equation gives us:

$\,\displaystyle \int^{v}_{27}dv=\int^{t}_{0}-6t\,dt$

$v\Big|^{v}_{27}=\dfrac{-6t^2}{2}\Big|^{t}_{0}$

$v-27=-3t^2$

$t=\sqrt{\dfrac{v-27}{-3}}$

When the sphere comes to a stop, v = 0. Thus, the time is:

$t=\sqrt{\dfrac{v-27}{-3}}$

$t=3$ s

To find the distance traveled, remember that:

$v=\dfrac{ds}{dt}$

$ds=v\,dt$

Rearranging the equation we found for the time so that velocity is isolated:

$v-27=-3t^2$

$v=-3t^2+27$

Substitute this value into our equation and take the integral:

$ds=v\,dt$

$\,\displaystyle \int^{s}_{0}ds=\int^{t}_{0}(-3t^2+27)dt$

$s=-t^3+27t$

When t = 3 seconds, we have:

$s=-(3)^3+27(3)$

$s=54$ m