Question

A car Starts from rest with an acceleration
of 4 m/s² ? What is the distance travelled in
8th Second ?​

Answer 1

EXPLANATION.

Car starts from rest

Acceleration of = 4m/s².

Distance travelled in 8th seconds.

As we know that,

Initial velocity = u = 0.

⇒ a = 4 m/s².

Formula of :

⇒ Sₙ = u + a/2(2n - 1).

⇒ S₈ = 0 + 4/2[2(8) - 1].

⇒ S₈ = 2[16 - 1].

⇒ S₈ = 2[15].

⇒ S₈ = 30 m.

Answer 2

Given :-

Acceleration = 8 m/s

To Find :-

Distance travelled in 8th second

Solution :-

We know that

Sₙ = u + a/2(2n - 1).

Here,

u = 0

a = 4

n = 8

By putting values

$\sf S_8 = 0 + \dfrac{4}{2} \bigg(2(8) - 1 \bigg)$

$\sf S_8 = 0 + 2\bigg(2(8) - 1\bigg)$

$\sf S_8 = 2 \times 15$

$\sf S_8 = 30 m$

Distance travelled is 30 m