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:

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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}}

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

t=3 s

To find the distance traveled, remember that:

v=\dfrac{ds}{dt}

ds=v\,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

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

\,\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

s=54 m

#### Final Answer:

The sphere travels 54 m.