Time Equal. Uniform Acceleration. Distance = ???

The initial speed of body will be equal to zero. So, in ${ t }_{ 1 }$ seconds, ${ x }_{ 1 }$ distance is traveled, in ${ t }_{ 2 }$ seconds, ${ x }_{ 2 }$ distance is traveled and in ${ t }_{ 3 }$ seconds, ${ x }_{ 3 }$ distance is traveled. Since $s=ut+\frac { 1 }{ 2 } a{ t }^{ 2 }$ , and when u=0 so in this case it reduces to $s=\frac { 1 }{ 2 } a{ t }^{ 2 }$

${ x }_{ 1 } = \frac { 1 }{ 2 } a{ { t }_{ 1 } }^{ 2 }$ $\Rightarrow$ ${ x }_{ 1 } = \frac { 1 }{ 2 } a{ 20 }^{ 2 }$ = 200a

${ x }_{ 2 }=\frac { 1 }{ 2 } a({ { t }_{ 2 } }^{ 2 }-{ { t }_{ 1 } }^{ 2 })$ $\Rightarrow$ ${ x }_{ 2 }=\frac { 1 }{ 2 } a({ 40 }^{ 2 }-{ 20 }^{ 2 })$ = 600a

${ x }_{ 3 }=\frac { 1 }{ 2 } a({ { t }_{ 3 } }^{ 2 }-{ { t }_{ 2 } }^{ 2 })$ $\Rightarrow$ ${ x }_{ 3 }=\frac { 1 }{ 2 } a({ 60 }^{ 2 }-{ 40 }^{ 2 })$ = 1000a

So, * A : B : C* :: 1 : 3 : 5

$\boxed{A+B+C = 9}$

As we know x is directly proprtional to the square of time taken. So distance in first part =x, distance in first part+second part=4x distance in first part+second part+third part=9x We directly conclude the ratio as 1:3:5 and answer will be 1+3+5=9

They are three distances then the answer is 3^2 = 9