Question

A particle moves along a straight line AB with constant acceleration. Its velocities are u

and v at A and B respectively. Show that its velocity at the mid-point of AB is ?​

Answer 1

Answer:

√v²+u² / 2

Step-by-step explanation:

Midpoint of AB  = C

Let distance AB = s       ,      then AC = s / 2  , velocity at c = vₓ

      In AB                                      In AC

v² = u² + 2as                        vₓ² = u² + 2as/2

v² - u² = 2as ---------{i}          vₓ² - u² = + as ---------{ii}  

from {i} and {ii}

vₓ² - u² = v² - u²/ 2

2(vₓ² - u²) = v² - u²

2vₓ² - 2u² = v² - u²

2vₓ² = v² - u²+ 2u²

2vₓ² = v² + u²

vₓ² = v² + u²/ 2

vₓ = √ v² + u²/ 2

∴velocity at the mid-point of AB is √ v² + u²/ 2

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