Question

The distance travelled by a particle starting from rest and.moving with an acceleration 4/3m/s^2 in the third second is?

Answer 1

here is your answer

Answer 2

Answer:

The distance covered in the third second is 3.33 m.

Explanation:

Given that,

Acceleration $a =\dfrac{4}{3}\ m/s^2$

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

Put the value of u and a in equation (I)

$s=0+\dfrac{1}{2}\times\dfrac{4}{3}t^2$

$s=\dfrac{2}{3}t^2$

We need to calculate the distance in the third second.

So, Firstly we need to calculate the distance in t = 2 s and t = 3 s.

For t = 2 s,

The distance is

$s_{1}=\dfrac{2}{3}\times(2)^2$

$s_{1}=2.67\ m$

For t = 3 s,

The distance is

$s_{2}=\dfrac{2}{3}\times(3)^2$

$s_{2}=6\ m$

Now, The distance is s

$s=s_{2}-s_{1}=6-2.67$

$s=3.33\ m$

Hence, The distance covered in the third second is 3.33 m.