Question

Derive the equation for uniform accelerated motion for the displacement covered in its nth second of its motion

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Answer 1

Answer:

Distance travelled in the nth second (Sn)=distance travelled in n (s1)seconds (sn)- distance travelled in n-1 second(s2)

          sn=s1-s2

distance travelled in t seconds s=ut+1/2at²

distance travelled in n seconds s1=un+1/2 an²

Distance travelled  in n-1 seconds s2=u(n-1) +1/2a(n-1)²

             sn=s1-s2

            =(un+1/2an²)-[u[n-1]+1/2a(n-1)²]

            =un+1/2an²-[un- u+1/2a(n2-2n+1)

            =un+1/2an²-un+u-1/2an²+an-1/22

            =u+an-1/2a

            sn=u+a[n-1/2]

(OR)

displacement from t= 0 to n sec.

 = s (n) = u n + 1/2 a n²

displacement from t = 0 to n-1 sec.

 = s(n-1) = u (n-1) + 1/2 a (n-1)²

 = u n - u + 1/2 a n² + 1/2 a - a n

Displacement in n'th second

 = Sn = s(n) - s(n-1)

          = u n + 1/2 a n² - u n + u - 1/2 a n² - 1/2 a + an

          = u + a (n - 1/2 )

Explanation:

Answer 2

ANSWER :

For getting the derivation , refer the attachment..

EXTRA INFO :

Motion :

If a body is changing it's position from one place to another place and with respect to time , the body is said to be in motion.

Rest :

If a body is not changing it's position with respect to time and place , then the body is said to be in rest.

Equations of uniform accelerated motion :

$v = u + at$

$s = ut + \frac{1}{2} a {t}^{2}$

${v}^{2} - {u}^{2} = 2as$

Equations for freely falling bodies :

$v = gt$

$h = \frac{1}{2} g {t}^{2}$

${v}^{2} = 2gh$

( If a = + g )

$h_{n} = g(n - \frac{1}{2} )$

Equations for vertically projected body :

$v = u - gt$

$h = ut - \frac{1}{2} g {t}^{2}$

${v}^{2} - {u}^{2} = - 2gh$

$h_{n} = u - g(n - \frac{1}{2} )$

NOTE :

When a body reaches the maximum height after projection , the final velocity becomes zero.

FINAL DERIVED FORMULA :

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