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

Question

Quizplus · Accepted Answer

We have given the velocity function $v ( t ) = \sin a t \cos ^ { 3 } a t$.
Using integration method, we can obtain the position function,
$$s(t)=\int v(t)dt$$
$$s(t)=-\frac{1}{4a}\cos^{4}(at)+C$$
As we are given that f(0) = 0, we can get the value of C.
$$0=-\frac{1}{4a}\cos^{4}(0)+C$$
$$C=\frac{1}{4a}$$
Therefore, the position function is given by
$$s(t)=-\frac{1}{4a}\cos^{4}(at)+\frac{1}{4a}$$
$$s(t)=\frac{1}{4a}\left ( 1-\cos^{4}(at) \right )$$
Hence, the answer is option D.

A Particle Moves on a Straight Line with Velocity Function v(t)=sin⁡atcos⁡3atv ( t ) = \sin a t \cos ^ { 3 } a tv(t)=sinatcos3at

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