Question

An object starts from rest recover a distance of 125m with acceleration of 10m/s find its final velocity and velocity 5 sec after starting journey

Answer 1

Answer:

${\bold{\therefore Final\:velocity=50\:m/s}}$

${\bold{\therefore Velocity\:after\:5\:sec=50\:m/s}}$

Explanation:

${\bold{\huge{\underline{SOLUTION-}}}}$

• In the given question information given about an object which is in rest and start acceleration of 10 m/s^2 and covers a certain distance.

• We have to find the final velocity and velocity after 4 sec.

$\underline \bold{Given : } \\\\ \implies Initial \: velocity (u) = 0\\ \\ \implies Distance(s) = 125 \: m \\ \\ \implies Acceleration(a) = 10 \: m/ {s}^{2} \\ \\ \underline \bold{To \: Find :} \\ \\ \implies Final \: velocity( v_{ 1 } ) = ? \\\\ \implies Velocity \: after \: 5 \: sec( v_{2} ) = ?$

• According to given question :

$\bold{By \: Second \: equation \: of \: motion : } \\\\ \implies {v}^{2} = {u}^{2} + 2as \\ \\\implies {v}^{2} = {0}^{2} + 2 \times 10 \times 125 \\ \\ \implies {v}^{2} = 2500 \\\\ \implies v = \sqrt{2500} \\ \\ \bold{\implies v = 50 \: m/s} \\ \\ \bold{for \: velocity \: afte r \: 5 \: sec : } \\ \\ \bold{By \: First \: equation \: of \: motion : } \\ \\\implies v = u + at \\ \\ \implies v_{ 2 } = 0 + 10 \times 5\\ \\ \bold{\implies v_{ 2 } = 50 \: m/s}$

Answer 2

Solution:

Given:

➜ An object starts from rest recover a distance of 125m with acceleration of 10m/s.

Find:

➜ Find its final velocity and velocity 5 sec after starting journey.

According to the given question:

➜ 0 = initial velocity.

➜ 125 m = distance.

➜ 10 m/s² = acceleration.

Know terms:

➜ Initial velocity = (u)

➜ Final velocity = (v)

➜ Acceleation = (a)

➜ Time = (t)

By using second equation of motion :

➜ v² = u² + 2as

➜ v² = 0² + 2 × 10 × 125

➜ v² = 2500

➜ v = √2500

➜ v = 50 m/s

By using second equation of motion:

➜ v = u + at

➜ v2 = 0 + 10 × 5

➜ v2 = 50 m/s

Therefore, v2 = 50 m/s is the required answer.

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