Position, Velocity and Acceleration
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A particle moves in a straight line so that the position \(x\,m\), relative to \(0\) at time \(t\) seconds \(t\ge0\) is given by \(x=t^2-6t+15\). Find
i) Its initial position.
ii) Its initial velocity.
iii) Its average velocity in the first \(4\)s
$$\begin{aligned}
\text{(i)}\quad &x=t^{2}-6 t+15 \\
&t=0, x=15 \mathrm{~m}\\
\text{(ii)}\quad &v=\frac{d x}{d t}=2 t-6\\
&t=0, \quad v=-6 \mathrm{~ms}^{-1}\\
&\text{Particle is moving to the left at}\ 6 ms^{-1}\\
&\begin{aligned}
\text{(iii)}\quad t=4 x &=4^{2}-6 \times 4+15 \\
&=16-24+15 \\
&=7 \\
v_{\alpha v} &=\frac{7-15}{4-0}=-\frac{8}{4}=-2 m {s}^{-1}
\end{aligned}\end{aligned}
$$
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