Question

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

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Answer 1

Explanation:

The distance travelled by a particle starting from rest and moving with an acceleration 43ms-2, in the third second is. Where u is initial speed and a is acceleration of the particle. ∴S3rd=0+12×43×(2×3-1)=103m.

Answer 2

Given :-

Acceleration of the particle = 4/3 m/s

Initial velocity of the particle = 0 m/s

To Find :-

Distance travelled by a particle in the third second.

Solution :-

We know that,

• a = Acceleration
• t = Time
• u = Initial velocity
• s =  Distance
• n = Integer

Using the formula,

$\underline{\boxed{\sf Second \ equation \ of \ motion=s=ut+\dfrac{1}{2} at^2}}$

Given that,

Acceleration (a) = 4/3 m/s

Initial velocity (u) = 0 m/s

Integer (n) = 3 sec

Substituting their values,

$\sf s=u+\dfrac{1}{2} a(2n-1)$

$\sf s=0+\dfrac{1}{2} \times \dfrac{4}{3} \times (2 \times 3-1)$

$\sf s=\dfrac{4}{6} \times 5$

$\sf s=\dfrac{10}{3} =3.33 \ m$

Therefore, the distance travelled in the third second is 3.33 m.