Jan Dangerfield, Stuart Haring and, Julian Gilbey Solutions for Exercise 3: EXERCISE 6A

At time $t$ s after jumping from a plane, the distance fallen by a parachutist is modelled as $s \mathrm{m}$, where
$s=\left\{\begin{array}{l}5t^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}:0⩽t⩽4\\ A\sqrt{t}+Bt\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}:4<t⩽25\\ Ct+30\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}:25<t⩽50\end{array}\right.$
$A, B$ and $C$ are constants. Explain why $A+2B=40$

At time $t$ s after jumping from a plane, the distance fallen by a parachutist is modelled as $s \mathrm{m}$, where
$s=\left\{\begin{array}{l}5t^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}:0⩽t⩽4\\ A\sqrt{t}+Bt\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}:4<t⩽25\\ Ct+30\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}:25<t⩽50\end{array}\right.$
$A, B$ and $C$ are constants. The parachute is opened at $t=4$ and the speed of the parachutist is immediately reduced by $x \mathrm{m}{ \mathrm{s}}^{-1}$. Show that $0.25 A+B=40-x$

At time $t$ s after jumping from a plane, the distance fallen by a parachutist is modelled as $s \mathrm{m}$, where
$s=\left\{\begin{array}{l}5t^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}:0⩽t⩽4\\ A\sqrt{t}+Bt\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}:4<t⩽25\\ Ct+30\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}:25<t⩽50\end{array}\right.$
$A, B$ and $C$ are constants.  At $t=25$ the speed of the parachute becomes constant. Write down two equations that connect $A, B$ and $C.$

At time $t$ s after jumping from a plane, the distance fallen by a parachutist is modeled as $s \mathrm{m}$, where
$s=\left\{\begin{array}{l}5t^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}:0⩽t⩽4\\ A\sqrt{t}+Bt\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}:4<t⩽25\\ Ct+30\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}:25<t⩽50\end{array}\right.$
Where $A, B$ and $C$ are constants. The parachute is opened at $t=4$ and the speed of the parachutist is immediately reduced by $x \mathrm{m}{ \mathrm{s}}^{-1}$.   At $t=25$ the speed of the parachute becomes constant. Find the value of $x.$ 

A particle moves forwards and backwards along a straight line. The displacement of the particle, $s \mathrm{m}$, from its initial position $O$ is given by $s=-0.6t^4+2.4t^3-3.6t^2+2.4t$ for $0<t<4$, where $t \mathrm{s}$ is the time for which the particle has been travelling.
Show that the particle starts moving along the line in the positive direction.

A particle moves forwards and backward along a straight line. The displacement of the particle, $s \mathrm{m}$, from its initial position $O$ is given by $s=-0.6t^4+2.4t^3-3.6t^2+2.4t$ for $0<t<4$, where $t \mathrm{s}$ is the time for which the particle has been travelling. The particle comes to instantaneous rest at point $A$, returns to pass through $O$ and continues to point $B$, where the distance $OB$ is the same as the distance $OA.$ Find the speed of the particle when it is at $B$. 

A robot moves along a straight line for $10 \mathrm{s}$. The displacement of the robot, $s \mathrm{m}$, from its initial position is given by $s=-0.01t^4+0.2t^3-1.32t^2+3.2t$, where $t$ is measured in seconds and $0<t<10.$ Show that the robot is stationary when $t=2, t=5$ and $t=8.$

A robot moves along a straight line for $10 \mathrm{s}$. The displacement of the robot, $s \mathrm{m}$, from its initial position is given by $s=-0.01t^4+0.2t^3-1.32t^2+3.2t$, where $t$ is measured in seconds and $0<t<10.$ The robot is stationary when $t=2, t=5$ and $t=8.$  Find the distance that the robot travels in the $10 \mathrm{s}$.