Question

#MODS challenge
Physics 2019 problem​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\frac{ v_{2} - v_{1} }{\Delta t_{1} + \Delta t_{2}} = \frac{v_{3} - v_{2}}{\Delta t_{3} + \Delta t_{2}}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Uniform \: acceleration = a \\ \\ \tt: \implies Successive \: time \: intervals = \Delta t_{1},\Delta t_{2} \: and \: \Delta t_{3} \\ \\ \tt: \implies Sucessive \: average \: velocity = v_{1}, v_{2} \: and \: v_{3} \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Correct \: relation =?$

• According to given question :

$\text{let \: initial \: velocity \: be \: u} \\ \\ \bold{As \: we \: know \: that} \\ \tt : \implies v = u + at \\ \\ \tt: \implies v \degree = u + a {\Delta t_{1} } - - - - - (1) \\ \\ \bold{Similarly : } \\ \\ \tt: \implies v \degree \degree = u + a(\Delta t_{1} + \Delta t_{2}) - - - - - (2) \\ \\ \bold{Again : } \\ \\ \tt: \implies v \degree \degree \degree = u + a(\Delta t_{1} +\Delta t_{2} + \Delta t_{3}) - - - - - (3) \\ \\ \bold{For \: first \: sucessive \: average} \\ \\ \tt: \implies v_{1} = \frac{u + v \degree}{2} \\ \\ \tt: \implies v_{1} = \frac{u + u +a {\Delta t_{1} } }{2} \\ \\ \tt: \implies v_{1} =u + \frac{a {\Delta t_{1} }}{2} - - - - - (4) \\ \\ \bold{Similarly \: for \: second \: sucessive}$

$\tt: \implies v_{2} = \frac{v \degree + v\degree \degree}{2} \\ \\ \tt: \implies v_{2} = \frac{ u + a {\Delta t_{1} } +u + a ({\Delta t_{1} + \Delta t_{2})}}{2} \\ \\ \tt: \implies v_{2} =u + a {\Delta t_{1} } + \frac{a\Delta t_{2}}{2} - - - - - (5) \\ \\ \bold{Similary : } \\ \\ \tt: \implies v_{3} = \frac{v \degree\degree +v \degree\degree \degree}{2} \\ \\ \tt: \implies v_{3} = \frac{u + a (\Delta t_{1} + \Delta t_{2}) +u + a (\Delta t_{1} + \Delta t_{2} + \Delta t_{3})}{2} \\ \\ \tt: \implies v_{3} = u + a\Delta t_{1} + a\Delta t_{2} + \frac{a\Delta t_{3} }{2} - - - - - (6) \\ \\ \text{Subtracting \: (4) \: from \: (5)} \\ \\ \tt: \implies v_{2} - v_{1} = u + a {\Delta t_{1} } + \frac{a\Delta t_{2}}{2} - u - \frac{a {\Delta t_{1} }}{2}$

$\tt: \implies v_{2} - v_{1} = \frac{a(\Delta t_{1} + \Delta t_{2})}{2} - - - - - (7) \\ \\ \text{Subtracting \: (5) \: from \: (6)} \\ \tt: \implies v_{3} - v_{2} = u + a {\Delta t_{1} } +a\Delta t_{2} + \frac{a\Delta t_{3}}{2} - u - a\Delta t_{1} - \frac{a {\Delta t_{2} }}{2} \\ \\ \tt: \implies v_{2} - v_{1} = \frac{a(\Delta t_{2} + \Delta t_{3})}{2} - - - - - (8) \\ \\ \text{Dividing \: (7) \:upon \: (8)} \\ \\ \tt: \implies \frac{ v_{2} - v_{1} }{ v_{3} - v_{2}} = \frac{ \frac{a(\Delta t_{1} + \Delta t_{2})}{2} }{\frac{a(\Delta t_{2} + \Delta t_{3})}{2}} \\ \\ \tt: \implies \frac{ v_{2} - v_{1} }{ v_{3} - v_{2}} =\frac{a(\Delta t_{1} + \Delta t_{2})}{2} \times \frac{2}{a(\Delta t_{2} + \Delta t_{3})} \\ \\ \tt: \implies \frac{ v_{2} - v_{1} }{ v_{3} - v_{2}} = \frac{(\Delta t_{1} + \Delta t_{2})}{(\Delta t_{2} + \Delta t_{3})} \\\\ \green{\tt:\implies \frac{ v_{2} - v_{1} }{\Delta t_{1} + \Delta t_{2}} = \frac{v_{3} - v_{2}}{\Delta t_{3} + \Delta t_{2}}}$