Question

the distance covered in 6th second is 32m. if the particle is moving with an initial velocity 10m/s.find its acceleration

Answer 1

Given :

▪ Initial velocity = 10mps

▪ Distance covered by particle in 6th second = 32m

To Find :

▪ Acceleration of the particle.

SoluTioN :

✴ Distance covered by moving body in nth second of journey is given by

$\bigstar\:\underline{\boxed{\bf{\red{s_n=u+\dfrac{a}{2}(2n-1)}}}}$

• s_n denotes distance covered in nth second
• u denotes initial velocity
• a denotes acceleration
• n denotes no. of second

$\Rightarrow\sf\:s_n=u+\dfrac{a}{2}(2n-1)\\ \\ \Rightarrow\sf\:32=10+\dfrac{a}{2}[2(6)-1]\\ \\ \Rightarrow\sf\:22=\dfrac{a}{2}(12-1)\\ \\ \Rightarrow\sf\:44=11a\\ \\ \Rightarrow\underline{\boxed{\bf{\purple{a=4\:ms^{-2}}}}}\:\orange{\bigstar}$

Answer 2

Given ,

" The distance covered in 6th second is 32 m. if the particle is moving with an initial velocity 10 m/s "

So ,

nth second  , n = 6 s

Distance , s = 32 m

Initial velocity , u = 10 m/s

We need to find the acceleration .

Acceleration , a = ? m/s

$\implies \bf s=u+ \dfrac{a}{2}(2n-1) \\\\\implies \bf 32=10+\dfrac{a}{2} (2(6)-1)\\\\\implies \bf 32=10+5.5a\\\\\implies \bf 5.5a=22\\\\\implies \bf a = 4\;m/s^2$