Review question

Can we find the velocity and acceleration of this particle? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8071

Solution

A particle $P$ moves in a straight line so that, after $t$ seconds, its distance, $s$ metres, from a fixed point $O$ is given by the equation. $s=12\sin\left(\frac{3}{2}t\right)$ Determine

1. expressions for the velocity and acceleration of $P$ in terms of $t$,
The velocity of the particle is the rate of change of its displacement from $O$. So we can differentiate the given function with respect to time. \begin{align*} v&=\frac{ds}{dt}=\frac{3}{2}\times12\cos\left(\frac{3}{2}t\right) \\ &=18\cos\left(\frac{3}{2}t\right) \end{align*} We can differentiate this function now to give us an expression for the acceleration. \begin{align*} a&=\frac{dv}{dt}=-\frac{3}{2}\times18\sin\left(\frac{3}{2}t\right) \\ &=-27\sin\left(\frac{3}{2}t\right) \end{align*}
1. the time at which $P$ first returns to $O$,

The graph of $s$ is a sine wave that has been stretched horizontally by a factor of $\frac{2}{3}$. A graph of $\sin t$ would meet the axis at $t=\pi$ so this graph will meet the axis at $t=\pi\times\frac{2}{3}$ The particle first returns to $0$ at $t=\frac{2}{3}\pi\approx\quantity{2.1}{s}$.

1. the maximum distance of $P$ from $O$ during the motion,

The graph has been stretched vertically so that it has an amplitude of $12$. The maximum distance from $O$ is equal to $\quantity{12}{m}$.

1. the velocity of $P$ when $s=6$.
We can substitute this value into the given equation giving us \begin{align*} 6&=12\sin\left(\frac{3}{2}t\right) \\ \implies\quad \sin\left(\frac{3}{2}t\right)&=\frac{1}{2} \end{align*}

Using this standard triangle we can see that if $\sin\left(\frac{3}{2}t\right)=\frac{1}{2}$ then $\cos\left(\frac{3}{2}t\right)=\frac{\sqrt{3}}{2}$. Therefore using the expression from part (1) we can calculate the velocity as $v=18\cos\left(\frac{3}{2}t\right)=18\frac{\sqrt{3}}{2}=9\sqrt{3}\approx\quantity{15.6}{m\,s^{-1}}.$