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Question

A particle with initial velocity $$v_0$$ moves with constant acceleration in a straight line. Find the distance travelled in $$n^{th}$$ second.


Solution

At t= n sec, distance traveled by object is
$${ s }_{ n }=u(n)+\dfrac { 1 }{ 2 } a{ (n) }^{ 2 }$$
In (n-1) sec, distance traveled by particle is
$${ s }_{ (n-1) }=u(n-1)+\dfrac { 1 }{ 2 } a{ (n-1) }^{ 2 }$$
Therefore distance covered in $${ n }^{ th }$$ sec, is
$${ s }_{ n }-{ s }_{ (n-1) }=un+\dfrac { 1 }{ 2 } a{ n }^{ 2 }-[u(n-1)+\dfrac { 1 }{ 2 } a{ (n-1) }^{ 2 }]\\ \quad \quad \quad \quad \quad \quad =\quad un+\dfrac { 1 }{ 2 } a{ n }^{ 2 }-[un-u+\dfrac { 1 }{ 2 } a({ n }^{ 2 }-2n+1)]$$
$$=u+an-\dfrac { 1 }{ 2 } a$$
$$\therefore \quad { s }_{ nth }=u+\dfrac { 1 }{ 2 } a[2n-1]$$


Physics

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