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NEET 2026 (Re-neet) · Distance, Displacement & Average Speed

Distance, Displacement & Average Speed question from NEET 2026 (Re-neet)

Q4NEET 2026 · RE-NEETMedium

Distance, Displacement & Average Speed

Consider a particle moving along a straight line, whose position as a function of time is given by s(t)=αt2−βt+γs(t) = \alpha t^2 - \beta t + \gamma, where α=1 ms−2,β=6 ms−1\alpha = 1\text{ ms}^{-2}, \beta = 6\text{ ms}^{-1} and γ=5 m\gamma = 5\text{ m}. The average speed of the particle, in ms−1\text{ms}^{-1} from t=0t = 0 to t=6 st = 6\text{ s} is:

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Answer: (d)

  1. s=t2−6t+5s = t^2 - 6t + 5, v=2t−6=0v = 2t - 6 = 0 at t=3t = 3 s, so the particle turns back.
  2. s(0)=5s(0) = 5, s(3)=−4s(3) = -4, s(6)=5s(6) = 5 m.
  3. Distance =∣−4−5∣+∣5−(−4)∣=9+9=18= |{-4} - 5| + |5 - (-4)| = 9 + 9 = 18 m
  4. Average speed =186=3= \frac{18}{6} = 3 m/s, so the answer is (d) 3.
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