Top speed

There are two distances covered Let X1 be the distance covered in 2 seconds And X2 be the distance covered in the 9 seconds left . X1=2a + 12 ; X2=9V But X1 + X2 = 100 =>2a + 12 +9V = 100 But a= (v-u)/t= (V-6)/2 => 2(V-6)/2 + 12 + 9V = 100 =>V-6+12+9V= 100 =>10V+6=100 =>10V=94 =>V=9.4 m/s

$Given \space distance \space of \space whole \space journey \space =\space 100 \space m \space and \space total \space time \space=\space 11 \space s.$

$First \space case \space : \space accelerated \space motion \space where \space initial \space velocity \space(u) \space =\space 6 \space m/s \space , \space acceleration \space(a)\space =\space 2 \space m/s^{2}. \space Let \space the \space final \space velocity \space be \space *v* \space m/s \space and \space time \space taken \space be \space *x* \space s.$

$Second \space case \space:\space uniform \space motion \space where \space the \space constant \space velocity \space =\space *v* \space and \space time \space taken \space is \space *(11-x)*.$

$Applying \space first \space eq \space of \space motion \space in \space First \space case \space: v \space = \space u \space + \space at$

$\rightarrow \space v \space=\space 6 \space+\space 2x$

$Applying \space second \space eq \space of \space motion \space in \space First \space case \space: s \space=\space u t \space+ \space \frac{1}{2}\space a t ^{2}$

$\rightarrow S_{1} \space=\space 6x \space + \space x^{2}$

$\rightarrow \space S_{2} \space =\space v\space(11-x) \space = \space (6\space + \space 2x)(11 \space - \space x)$

$Here \space,\space S_{1} \space and \space S_{2} \space are \space the \space distances \space travelled \space in \space first \space and \space second \space cases \space respectively$

$\rightarrow \space S_{1} \space + \space S_{2} \space = \space 100$

$\rightarrow \space 6x \space+ \space x^{2} \space + \space (6 \space+\space 2x)(11 \space-\space x) \space = \space 100$

$Solving \space for \space x \space , \space we \space get \space x \space=\space (11 \space + \space \sqrt{87} )\space,\space (11\space - \space \sqrt{87})$

$As \space total \space time \space = \space 11 \space s \space \therefore \space x \space cannot \space be \space greater \space than \space 11 \space \therefore \boxed{x \space = \space 11 \space - \space \sqrt{87}}$

$V_{max} \space = \space *v* \space=\space 6 \space + \space 2x$

$\Rightarrow \space V_{max} \space=\space 9.346\space \sim \space 9.4 \space m/s$