Question

A car starting from rest is accelerated uniformly at the rate of 2 m/s^2. Determine the velocity of the car when it covers a distance of 50 m.
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Answer 1

Provided that:

• Initial velocity = 0 m/s
• Acceleration = 2 m/s sq.
• Distance = 50 m

To calculate:

• The final velocity

Solution:

• The final velocity = 14.14 m/s

Using concept:

• Third equation of motion

Using formula:

• v² - u² = 2as

Required solution:

→ v² - u² = 2as

→ v² - (0)² = 2(2)(50)

→ v² - (0)² = 2(100)

→ v² - 0 = 200

→ v² = 200

→ v = √200

→ v = 14.14 m/s

→ Final velocity = 14.14 m/s $\:$

Answer 2

Answer:

14.14 m/s^2

Explanation:

We have to solve this from 3rd Equation of Motion i.e. $2as = v^{2}-u^{2}$

We know that car started from rest so its initial velocity will be 0 m/s.

So,

• Initial Velocity = 0 m/s

• Acceleration = 2 m/s^2

• Distance = 50 m  

  So,  $2as = v^{2}-u^{2}$

          2×2×50 = $v^{2} - 0^{2}$

          200 = $v^{2}$

          $v=\sqrt{200}$

         $v =$ 14.14 m/s

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