Can we find the velocity and acceleration of this particle?

Can we find the velocity and acceleration of this particle?

Question

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 $t$,

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1. the time at which $P$ first returns to $O$,

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

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1. the velocity of $P$ when $s=6$.

Solution

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.

$$\begin{align*} a&=\frac{dv}{dt}=-\frac{3}{2}\times18\sin\left(\frac{3}{2}t\right) \\ &=-27\sin\left(\frac{3}{2}t\right) \end{align*}$$

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 $t=\frac{2}{3}\pi\approx2.1\,\text{s}$.

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

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 $v=18\cos\left(\frac{3}{2}t\right)=18\frac{\sqrt{3}}{2}=9\sqrt{3}\approx15.6\,{\mathrm{m\,s^{-1}}}.$