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JEE Main 2023 (13 Apr, Shift 2) · Velocity, Acceleration & Variable Motion

Velocity, Acceleration & Variable Motion question from JEE Main 2023 (13 Apr, Shift 2)

Q19JEE Main 2023 · 13 APR · SHIFT 2Easy

Velocity, Acceleration & Variable Motion

The distance travelled by an object in time tt is given by s=(2.5)t2s = (2.5)t^2. The instantaneous speed of the object at t=5 st = 5\text{ s} will be:

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Answer: (d)

  1. v=dsdt=2×2.5 t=5tv = \frac{ds}{dt} = 2 \times 2.5\, t = 5t
  2. At t=5t = 5 s: v=25v = 25 m/s
  3. So the answer is (d) 25 ms−125\text{ ms}^{-1}.
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