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A particle is travelling with a uniform acceleration. if $a$ , $b$ and $c$ are the distances covered by it during $x^{th}$ , $y^{th}$ and $z^{th}$ second of its motion respectively then

Option 1)
$ax = by = cz$

Option 2)
$a(y-z) + b(z-x) + c(x-y) = 0$

Option 3)
$ax + by = cz$

Option 4)
None of these.

Answers (1)

best_answer

As we have learnt,

Displacement in nth second -

$S_{n}= u+\frac{a}{2}(2n-1)$

- wherein

$u=$ Initial velocity

$a=$ uniform acceleration

n= n^th second

Distance travelled in n^thsecond is

$s_n = u + \frac{1}{2}A(2n-1)$

Here :

$a = u + \frac{1}{2}(2x -1)A$

$b = u + \frac{1}{2}(2y -1)A$

$c = u + \frac{1}{2}(2z -1)A$

or

$a = Ax + \left (u-\frac{A}{2} \right ) \\*\Rightarrow ay = Axy + \left(u -\frac{A}{2}\right )y \; \& \; az = Axz + \left(u -\frac{A}{2}\right )z$ \

Subtracting we get,

$a(y-z)= A(xy-xz) + (u - \frac{A}{2})\cdot (y-z)$

Similarly,

$b(z-x)= A(yz-yx) + (u - \frac{A}{2})\cdot (z-x)$

$c(x-y)= A(zx-yz) + (u - \frac{A}{2})\cdot (x-y)$

Adding these three equations, we get

$a(y-z) + b(z-x) + c(x-y) = 0$

Option 1)

$ax = by = cz$

Option 2)

$a(y-z) + b(z-x) + c(x-y) = 0$

Option 3)

$ax + by = cz$

Option 4)

None of these.

Posted by

Avinash

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