A cyclist rides in a straight line with constant decelaration $\\var{-{a}} \\mathrm{ms^{-2}}$ from a point $A$ to a point $B$. At point $A$ the cyclist's speed is $\\var{u}\\mathrm{ms^{-1}}$ and at point $B$ the speed is $\\var{v}\\mathrm{ms^{-1}}$.
", "advice": "a) You are given $u = \\var{u},$ $v=\\var{v}$, $a = \\var{a}$, $t = ?$, $s = ?$
\nNote that $a$ is negative because the cyclist is decelerating.
\nYou need $t$, so you use $v=u+at$, rearranged for $t$.
\n\\begin{align} t & = \\frac{v-u}{a}, \\\\
& = \\frac{\\var{v}-\\var{u}}{\\var{a}}, \\\\
& = \\var[fractionNumbers]{(v-u)/a} \\mathrm{s}. \\end{align}
The time taken to move from $A$ to $B$ is $\\var[fractionNumbers]{(v-u)/a}$ seconds.
\nb) You need the distance, $s \\mathrm{m}$. Therefore you can use $s = \\left(\\frac{u+v}{2}\\right)t$, with the $t$ value found in part a).
\n\\begin{align} s & = \\left(\\frac{u+v}{2}\\right)t, \\\\
& = \\left(\\frac{\\var{u}+\\var{v}}{2}\\right) \\times \\var[fractionNumbers]{(v-u)/a}, \\\\
& = \\var[fractionNumbers]{(u+v)*0.5*((v-u)/a)} \\mathrm{m}. \\end{align}
The distance from $A$ to $B$ is $\\var[fractionNumbers]{(u+v)*0.5*((v-u)/a)}$ metres.
c) You now have $u=\\var{u}$, $a = \\var{a}$, $t=\\var{t2}$, $s=?$ and you are trying to find the velocity at point $C$ so $v=?$
\nTherefore you can use $v = u+at$.
\n\\begin{align} v & = u + at, \\\\
& = \\var{u} + \\left(\\var{a} \\times \\var{t2}\\right), \\\\
& = \\var[fractionNumbers]{u+a*t2} \\mathrm{ms^{-1}}. \\end{align}
Note that the velocity can be negative - this means the cyclist is moving from right to left. Therefore the velocity of the cyclist at point $C$ is $\\var{-(u+a*t2)} \\mathrm{ms^{-1}}$ in the direction $\\simplify{vec:BA}.$
\nd) We now need the distance, $s \\mathrm{m}$. We have $u=\\var{u}$, $t=\\var{t2}$ and we have just evaluated the final velocity at point C, $v=\\var{u+a*t2}$.
\n\\begin{align} s & = \\left(\\frac{u + v}{2}\\right)t, \\\\
& = \\left(\\frac{\\var{u} + \\var{u+a*t2}}{2}\\right) \\times \\var{t2}, \\\\
& = \\var[fractionNumbers]{(u+u+a*t2)*0.5*t2} \\mathrm{m}. \\end{align}
The distance from A to C is $\\var[fractionNumbers]{(u+u+a*t2)*0.5*t2}$ metres.
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