Question

A particle starts at the origin with initial velocity \({\rm{i - j + 3k}}\).It’s acceleration is,\({\rm{a(t) = 6ti + 12}}{{\rm{t}}^{\rm{2}}}{\rm{j - 6tk}}\).Find its position function. Exercise \({\rm{54}}\)section \({\rm{10}}\).

Short answer

Therefore, the positional function of the particle is,

\(r(t) = \left( {{t^3} + t} \right)i + \left( {{t^4} - t} \right)j + \left( {3t - {t^3}} \right)k\).

Step-by-step solution

Defination of the positional function of the particle

We just multiply the velocity function to obtain the displacement (position shift). The overall displacement is decreased by the negative regions below the x-axis. We must use absolute value to compute the distance travelled.

Evaluating the Velocity of the particle

We know that,\(v(t) = \smallint a(t)dt\)

Given that, \(a(t) = 6ti + 12{t^2}j - 6tk\)

Therefore, \(v(t) = \smallint 6ti + 12{t^2}j - 6tkdt\)

\(v(t) = 3{t^2}i + 4{t^3}j - 3{t^2}k + C\)

Where C is a constant vector.

Given that, the Intial velocity of the particle is i-j+3k

Substitude t=0 in the Equation, to get

\(i - j + 3k = 3{(0)^2}i + 4{(0)^3}j - 3{(0)^2}k + C\)

Therefore,

\(i - j + 3k = C\)

Substitude the value of the C, to get

\(v(t) = 3{t^2}i + 4{t^3}j - 3{t^2}k + (i - j + 3k)\)

\(v(t) = \left( {3{t^2} + 1} \right)i + \left( {4{t^3} - 1} \right)j + \left( {3 - 3{t^2}} \right)k\)

The value of the positional function

We know that, \(r(t) = \smallint v(t)dt\)

Therefore,

\(\begin{array}{*{20}{c}}{r(t) = \smallint \left( {3{t^2} + 1} \right)i + \left( {4{t^3} - 1} \right)j + \left( {3 - 3{t^2}} \right)kdt}\\{r(t) = \left( {{t^3} + t} \right)i + \left( {{t^4} - t} \right)j + \left( {3t - {t^3}} \right)k + C}\end{array}\)

Where C is constant vector.

Given that the initial position of the particle is origin.

Substitude t=0 in the Equation, to get

\(0 = \left( {{0^3} + 0} \right)i + \left( {{0^4} - 0} \right)j + \left( {3(0) - {{(0)}^3}} \right)k + C\)

Therefore, 0=C

Substitude the value of C, to get

\(r(t) = \left( {{t^3} + t} \right)i + \left( {{t^4} - t} \right)j + \left( {3t - {t^3}} \right)k\)

Therefore, the positional function of the particle is,

\(r(t) = \left( {{t^3} + t} \right)i + \left( {{t^4} - t} \right)j + \left( {3t - {t^3}} \right)k\).