Question

its very hard question plz plz solve this. answer:
$\alpha \beta t \div 2( \alpha + \beta )$
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Answer 1

$\huge\fcolorbox{red}{pink}{Answer} \\ ............................................................\\ \implies \rm \: for \: A \: to \: B \\ \\ \leadsto \rm \: v = u + at \\ \\ \leadsto \rm \: v_{max} = 0 + \alpha{t_1} \\ \\ \leadsto \rm \: \blue{t_1 = \frac{v_{max}}{ \alpha}} \\ ............................................................ \\ \implies \rm \: for \: B \: to \: C \\ \\ \leadsto \rm \: v = u + at \\ \\ \leadsto \rm \: 0 = v_{max} - \beta{t_2} \\ \\ \leadsto \rm \: \green{t_2 = \frac{v_{max}}{ \beta}} \\ ............................................................ \\ \implies \rm \: total \: time \: is \: given \: by \\ \\ \leadsto \rm \: t = t_1 + t_2 \\ \\ \leadsto \rm \: t = \frac{v_{max}}{ \alpha} + \frac{v_{max}}{ \beta} \\ \\ \leadsto \rm \: \red{t = v_{max} \: ( \frac{ \alpha \beta}{ \alpha + \beta}) } \\............................................................ \\ \implies \rm \: total \: distance \: is \: given \: by \\ \\ \leadsto \rm \: d = ut + \frac{1}{2} a {t}^{2} \\ \\ \leadsto \rm \: d = 0 + \frac{1}{2} \times \frac{v_{max}} {t} \times {t}^{2} \\ \\ \leadsto \rm \: \orange{ d = \frac{1}{2} \times v_{max} \times t} \\ ............................................................\\ \implies \rm \: average \: velocity \: is \: given \: by \\ \\ \leadsto \rm \: v_{av} = \frac{total \: distance}{total \: time} \\ \\ \leadsto \rm \: v_{av} = \frac{ \frac{1}{2} \times v_{max} \times t \times ( \alpha + \beta)}{v_{max} \times \alpha \beta} \\ ............................................................\\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \pink{v_{av} = \frac{( \alpha + \beta)t}{ \alpha \beta}}}}}}$