Question

if sn=2+0.4n find initial velocity and acceleration

Answer 1

Given,
Sn = 2 + 0.4n
Here Sn is distance travelled in nth second.
We have to resolve this equation,
Sn = 2 + 0.4(2n -1)/2 + 0.4/2
Sn = 2.2 + 0.4(2n -1)/2
Sn = 2.2 + (2n -1)/2 (0.4)

we know, the formula of Sn = u + (2n -1)/2 a
Where a is Acceleration and u is initial velocity of object.
Compare both equations ,
u = 2.2 m/s and a = 0.4 m/s²

Hence, initial velocity = 2.2 m/s and acceleration = 0.4 m/s²

Answer 2

"The initial velocity is $2.2 m s^{-1}$ and the acceleration is $0.4 m s^{-2}$

Given:

$Sn = 2+0.4n \rightarrow (1)$

Initial velocity, u = ?

Acceleration, a = ?

Solution:

We know that the distance travelled in nth second is given by

$S_{n}=\left(u-\frac{1}{2} a\right)+a n \rightarrow(2)$

On substituting, we get

On comparing equations (1) and (2)

$a\quad =\quad 0.4{ ms }^{ -2 }$

$u-\left( \frac { a }{ 2 } \right) \quad =\quad 2$

$u-\left( \frac { 0.4 }{ 2 } \right) \quad =\quad 2$

$u\quad =\quad 2.2ms^{ -1 }$"