Question

A body moving with uniform acceleration describes 4 m in third second and 12 m in fifth second. Find the distance described in next three second?​

Answer 1

Solution :-

The distance covered by the body in Sn th second is given by ,

$\dag \:\boxed{\mathtt{S_n= u + \dfrac{a}{2}(2n-1)}}$

Third second

• Distance covered = 4 m

Now ,

$\longrightarrow\mathtt{S_3= u + \dfrac{a}{2}(2n-1)}$

$\longrightarrow\mathtt{4= u + \dfrac{a}{2}(2x3-1)}$

$\longrightarrow\mathtt{4= u + \dfrac{5a}{2}}$

$\longrightarrow\mathtt{8= 2u + 5a} ....(1)$

Fifth second

• Distance covered = 12 m

Now ,

$\longrightarrow\mathtt{S_5= u + \dfrac{a}{2}(2n-1)}$

$\longrightarrow\mathtt{12 = u + \dfrac{a}{2}(2x5-1)}$

$\longrightarrow\mathtt{12= u + \dfrac{9a}{2}}$

$\longrightarrow\mathtt{24= 2u + 9a} ....(2)$

Substracting (1) from (2)

$\longrightarrow\mathtt{16= 4a}$

$\longrightarrow\boxed{\mathtt{a = 4 \dfrac{m}{s^{2} } }}$

Substituting the value of a in (1) we get ,

$\longrightarrow\mathtt{8= 2u + 5x 4 }$

$\longrightarrow\mathtt{8= 2u + 20}$

$\longrightarrow\mathtt{2u = -12 }$

$\longrightarrow\boxed{\mathtt{u = -6 \dfrac{m}{s} }}$

Distance covered in 5 seconds

$\longrightarrow\mathtt{S = ut + \dfrac{1}{2}at^{2} }$

$\longrightarrow\mathtt{S= -6 x 5+ \dfrac{1}{2}(4x25)}$

$\longrightarrow\mathtt{S= - 30+ 50 }$

$\longrightarrow\mathtt{S= 20 m }$

Distance covered in 8 seconds

$\longrightarrow\mathtt{S'= ut + \dfrac{1}{2}at^{2} }$

$\longrightarrow\mathtt{S'= -6 x 8+ \dfrac{1}{2}(4x64)}$

$\longrightarrow\mathtt{S'= - 48+ 128 }$

$\longrightarrow\mathtt{S'= 80 m }$

Distance covered in next 3 second

$\longrightarrow\mathtt{S' - S }$

$\longrightarrow\mathtt{80 - 60 }$

$\longrightarrow\mathtt{60 m }$

The distance covered by the body in next 3 seconds is 60 m