Question

a body starts from rest and moves with a constant acceleration of 4 m/s . calculate the velocity acquired by it after it has covered a distance of 8m.​

Answer 1

▶ Solution :

Given :

▪ Initial velocity of body = zero (i.e. rest)

▪ Acceleration of body = 4m/s^2

To Find :

▪ Velocity acquired by it after it has covered a distance of 8m.

Concept :

✒ Since, acceleration is constant through out the motion we can apply equation of kinematics directly.

✒ Second equation of kinematics is given by

☞ v^2 - u^2 = 2as

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance travelled

Calculation :

→ v^2 - u^2 = 2as

→ v^2 - (0)^2 = 2(4)(8)

→ v^2 = 2×32

→ v^2 = 64

→ v = √(64)

→ v = 8mps

Answer 2

GiveN :

Initial velocity (u) = 0 m/s

Acceleration (a) = 4 m/s

Distance Traveled = 8m

To FinD :

Velocity when it have traveled 8 m

SolutioN :

We are given initial velocity, acceleration and also the distance traveled. So, we can calculate the value of velocity when body has covered 8m. As acceleration is constant, so we can use Kinematics Equations.

$\underbrace{\sf{Velocity \: when \: body \: has \: travelled \: 8 \: m}}$

Use 3rd equation of Kinematics

$\implies \sf{v^2 \: - \: u^2 \: = \: 2as} \\ \\ \implies \sf{v^2 \: - \: 0^2 \: = \: 2 \: \times \: 4 \: \times \: 8} \\ \\ \implies \sf{v^2 \: = \: 2(32)} \\ \\ \implies \sf{v^2 \: = \: 64} \\ \\ \implies \sf{v \: = \: \sqrt{64}} \\ \\ \implies \sf{v \: = \: \sqrt{8^2}} \\ \\ \implies \sf{v \: = \: 8} \\ \\ \underline{\sf{\therefore \: Velocity \: when \: body \: has \: travelled \: 8 \: m \: is \: 8 \: ms^{-1}}}$

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