Question

A body moving with uniform acceleration has velocities 20 m s-1 and 30 m s-1 wh
en passing two points A and B. Then the velocity midway between A and B is :​

Answer 1

$\color{red}\huge{\underline{\underline{\bf GIVEN\dag}}}$

• $\sf{velocity(u) = 20 {ms}^{ - 1} }$
• $\sf{velocity(v) = 30 {ms}^{ - 1} }$

$\color{red}\huge{\underline{\underline{\bf SOLUTION\dag}}}$

$\purple{\sf{let;}}$

$\sf acceleration(a) =a$

$\sf Displacement(s) = AB = x$

$\purple{\sf{we \: know \: that;}}$

$\red{ \boxed{ {v}^{2} - {u}^{2} = 2as }}$

$\implies \sf{{(30)}^{2} - {(20)}^{2} = 2ax}$

$\implies \sf{900 - 400 = 2ax}$

$\implies \sf{2ax = 500}$

$\implies \sf{a = \frac{250}{x} }$

$\purple{ \sf{at \: midpiont : }}$

$\sf{dispacement(s) = \frac{x}{2} }$

$\sf{velocity \: at \: mid \: point = v_{m} }$

$\purple{ \sf{applying \: formula : }}$

$\implies {(v_{m} )}^{2} - {(20)}^{2} = 2( \frac{250}{x} )( \frac{x}{2} )$

$\implies {(v_{m} )}^{2} = 250 + 400$

$\implies {(v_{m} )}^{2} = 650$

$\implies v_{m} = \sqrt{650}$

$\orange{ \therefore v_{m} = 25.5 {ms}^{ - 1} }$