A particle moves on a straight line with velocity function $v ( t ) = \sin \omega t \cos ^ { 3 } a t$. Find its position function $s = f ( t )$ if $f ( 0 ) = 0$.
A) $\frac { \cos ^ { 5 } \omega t + 1 } { 5 \omega }$
B) $\frac { \sin ^ { 2 } \omega t + 1 } { 7 a }$
C) $\frac { \sin ^ { 3 } \omega t - 1 } { 3 \omega }$
D) $\frac { 1 - \cos ^ { 4 } a t } { 4 a }$
E) $\frac { 1 - \cos ^ { 6 } a t + 1 } { 6 a }$

Question

Quizplus · Accepted Answer

Since velocity is the derivative of the position function, we can find the position function by integrating the given velocity function.

\begin{aligned} v(t) &= \sin \omega t \cos^3 at \ s(t) &= \int v(t) dt = \int \sin \omega t \cos^3 at \ dt \end{aligned}

We use the substitution u = sin(at) and du = acos(at)dt to simplify the integral.

\begin{aligned} s(t) &= \int \sin \omega t \cos^3 at \ dt \ &= - \int \sin \omega t \cdot (- \cos^3 at) \ dt \ &= - \int \sin \omega t \cdot (1-u^2) \cdot \frac {du} {a} \ &= - \frac {1} {a} \int \sin \omega t \cdot du + \frac {1} {a} \int \sin \omega t \cdot u^2 \cdot du \ &= - \frac {1} {a} \cos ( \omega t ) \cdot u - \frac {1} {a} \int ( \frac {1 - \cos 2 \omega t} {2} ) \cdot u^2 \cdot du \ &= - \frac {1} {a} \cos ( \omega t ) \cdot \sin ( a t ) - \frac {1} {2a} [ \int u^2 \cdot du - \int u^2 \cdot \cos 2 \omega t \cdot du ] \ &= - \frac {1} {a} \cos ( \omega t ) \cdot \sin ( a t ) - \frac {u^3} {6a} + \frac {u^3 \cdot \cos 2 \omega t} {6a} + C \ &= - \frac {1} {a} \cos ( \omega t ) \cdot \sin ( a t ) - \frac {\sin^3 at} {6a} + \frac {\sin^3 at \cdot \cos 2 \omega t} {6a} + C \end{aligned}

Now we substitute u = sin(at) and u(0) = 0 to get

s(t) = - (1/a) cos(wt)sin(at) - (1/(6a))sin^3(at) + (1/(6a))sin^3(at)cos(2wt) + C

Since s(0) = 0, we have C = 0. Thus,

s(t) = - (1/a) cos(wt)sin(at) - (1/(6a))sin^3(at) + (1/(6a))sin^3(at)cos(2wt)

Simplifying gives:

s(t) = (1/(4a))(1 - cos^4(at))

Therefore, the answer is D.

A Particle Moves on a Straight Line with Velocity Function v(t)=sin⁡ωtcos⁡3atv ( t ) = \sin \omega t \cos ^ { 3 } a tv(t)=sinωtcos3at

A Particle Moves on a Straight Line with Velocity Function $v ( t ) = \sin \omega t \cos ^ { 3 } a t$