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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.

Step-by-step

Velocity: { v }_{ 0 } = 27 m/s \text{ at } { t }_{ 0 } = 0 s. We have

\begin{aligned} (+ \downarrow) \quad\quad & dv = adt \\ & \int_{ 27 }^{ v } dv = \int_{ 0 }^{ t } -6tdt \\ & v = (27 – 3{ t }^{ 2 }) m/s \quad\quad\quad\quad (1) \end{aligned}

At v = 0 , from Eq. (1)

0 = 27 – 3{ t }^{ 2 }\quad\quad t = 3.00s

Distance Traveled: { s }_{ 0 } = 0 m at { t }_{ 0 } = 0s . Using the result v = 27 – 3{ t }^{ 2 }, we have

\begin{aligned} (+ \downarrow) \quad\quad & ds = vdt \\ & \int_{ 0 }^{ s } ds = \int_{ 0 }^{ t } (27 – 3{ t }^{ 2 })dt \\ & s = (27t – { t }^{ 3 }) m \quad\quad\quad\quad (2) \end{aligned}

At t = 3.00 s , from Eq. (2)

s = 27(3.00) – { 3.00 }^{ 3 } = 54.0m

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