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A body starts from rest and moves with uniform acceleration. If its velocity after 5 seconds is 20 m/s, what is the acceleration and the distance covered in 5 seconds?

Physics

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Answer

4 m/s2\text{4 m/s}^2

50 m\text{50 m}

Step-by-step solution

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The image provided is very difficult to read due to its low quality and blurriness. Based on the visible words and common problem structures, I will assume the question is:

"A body starts from rest and moves with uniform acceleration. If its velocity after 5 seconds is 20 m/s, what is the acceleration and the distance covered in 5 seconds?"

Here's the solution based on this assumption:

Given:

  • Initial velocity (uu) = 0 m/s (starts from rest)
  • Time (tt) = 5 s
  • Final velocity (vv) = 20 m/s

We need to find:

  • Acceleration (aa)
  • Distance covered (ss)
  1. Calculate the acceleration (aa).

    We use the first equation of motion: v=u+atv = u + at.

    20 m/s=0 m/s+a×5 s20 \text{ m/s} = 0 \text{ m/s} + a \times 5 \text{ s} 20=5a20 = 5a a=205a = \frac{20}{5} a=4 m/s2a = 4 \text{ m/s}^2

    The acceleration is 4 m/s2\boxed{\text{4 m/s}^2}.

  2. Calculate the distance covered (ss).

    We use the second equation of motion: s=ut+12at2s = ut + \frac{1}{2}at^2.

    s=(0 m/s)(5 s)+12(4 m/s2)(5 s)2s = (0 \text{ m/s})(5 \text{ s}) + \frac{1}{2}(4 \text{ m/s}^2)(5 \text{ s})^2 s=0+12(4)(25)s = 0 + \frac{1}{2}(4)(25) s=2×25s = 2 \times 25 s=50 ms = 50 \text{ m}

    The distance covered is 50 m\boxed{\text{50 m}}.

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