Question

A particle starts from rest with constant acceleration
= 2m/s2. Find displacement in 5th sec.
(A)9m
(B) 18m
(C) 15m
(D) 6m

Answer 1

Given :

• Acceleration = 2 m/s²
• nth second = 5th s.

To find :

The displacement in the nth second.

Solution :

We know the Equation for Distance traveled in nth second i.e,

$\bf{S = u + \dfrac{1}{2}a(2n - 1)}$

Where :

• S = Displacement
• u = Initial Velocity
• n = nth second
• a = Acceleration

Using the above equation and substituting the values in it, we get :

$:\implies \bf{S_{n} = u + \dfrac{1}{2}a(2n - 1)}$

$:\implies \bf{S_{5} = u + \dfrac{1}{2}a(2n - 1)}$

[Note : Here , the initial velocity will be 0 as the particle is starting from rest]

$:\implies \bf{S_{5} = 0 + \dfrac{1}{2}a(2n - 1)}$

$:\implies \bf{S_{5} = \dfrac{1}{2} \times 2 \times [2 \times (5) - 1]}$

$:\implies \bf{S_{5} = \dfrac{1}{2} \times 2 \times (10 - 1)}$

$:\implies \bf{S_{5} = \dfrac{1}{2} \times 2 \times 9}$

$:\implies \bf{S_{5} = \dfrac{1}{\not{2}} \times \not{}2 \times 9}$

$:\implies \bf{S_{5} = 9}$

$\boxed{\therefore \bf{S_{5} = 9\:m}}$

Hence the displacement in 5th is 9 m.

Answer 2

Answer:

9

Explanation: