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202408211136 Exercise 14.C.10

A stone is thrown vertically upwards so that its height, s metres, above the ground, after t seconds is given by s=40t-5t^2. Find

(a) its velocity after 2\,\mathrm{s};
(b) its height above the ground when it is momentarily at rest;
(c) its initial velocity;
(d) its velocity when it is 15\,\mathrm{m} above the ground, giving your answer correct to the nearest \mathrm{m\,s^{-1}}.

Extracted from K. S. Teh & C. Y. Loh. (2007). New Syllabus Additional Mathematics (8e)

Roughwork.

Initially (i.e., at t=0), the stone is

s(t=0)=40(0)-5(0)^2=0\,\mathrm{m}

above the ground, i.e., at ground level s=0.

\begin{aligned} & s = 40t-5t^2 = t(40-5t) \\ \Rightarrow & \begin{cases} s =0 \Leftrightarrow t = \{0\}\cup\{ 8\} \\ s > 0 \Leftrightarrow t\in (0,8) \\ s < 0 \Leftrightarrow t<0\enspace\textrm{(rej.)}\enspace\textrm{\scriptsize{OR}}\enspace t>8 \\ \end{cases} \\ \end{aligned}

Take upward positive. The (vertical) displacement of the stone is

\begin{aligned} \mathbf{s}(t) & =s(t)\,\hat{\mathbf{j}} \\ & = (40t-5t^2)\,\hat{\mathbf{j}} \\ \end{aligned}

its (vertical) instantaneous velocity

\begin{aligned} \mathbf{v}(t) & = v(t)\,\hat{\mathbf{j}} \\ & = \frac{\mathrm{d}s(t)}{\mathrm{d}t}\,\hat{\mathbf{j}} \\ & = (40t-5t^2)'\,\hat{\mathbf{j}} \\ & = (40-10t)\,\hat{\mathbf{j}} \\ \end{aligned}

and its (vertical) instantaneous acceleration

\begin{aligned} \mathbf{a}(t) & = a(t)\,\hat{\mathbf{j}} \\ & = \frac{\mathrm{d}v(t)}{\mathrm{d}t}\,\hat{\mathbf{j}} \\ & = (40-10t)'\,\hat{\mathbf{j}} \\ \mathbf{a} & = -10\,\hat{\mathbf{j}} \\ \end{aligned}

so are the answers to

(a) \mathbf{v}(2)
(b) t'\in\{ t:v(t)=0\}\enspace\rightsquigarrow\enspace s(t')
(c) \mathbf{v}(0)
(d) t'\in\{ t:\mathbf{s}(t)=+15\,\hat{\mathbf{j}}\} \enspace\rightsquigarrow\enspace \mathbf{v}(t')

This problem is not to be attempted.