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.