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  • An object accelerates from rest to velocity 27.5 m per sec in 10 sec. Find distance covered by object next 10 seconds

An object accelerates from rest to a velocity of 27.5 m/s in 10 seconds. Find the distance covered by the object in the next 10 seconds

Answers (1)

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The given data-

  • Initial velocity (u) = 0 m/s (since the object starts from rest)
  • Final velocity after first 10 seconds (v) = 27.5 m/s
  • Time (t) = 10 s

First to find the acceleration, the formula used is $$
v = u + at
$$

Putting the values in the formula, $$
27.5 = 0 + a(10)
$$
$$
a = \frac{27.5}{10} = 2.75 \text{ m/s}^2
$$

Now, calculate the distance covered in the next 10 seconds. At the start of the next 10 seconds, the object has an initial velocity of 27.5 m/s, and it continues with the same acceleration (2.75 m/s2). Using the second equation of motion :$$
s = ut + \frac12 at^2
$$

Putting the values in the formula, $$
s = (27.5 \times 10) + \frac{1}{2} (2.75) (10)^2 = 275 + 137.5 = 412.5 \text{ m}
$$

Therefore, the distance covered by the object in the next 10 seconds is 412.5 meters.

Posted by

Saniya Khatri

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