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A particle travels along a straight line from a point A on the line. Its velocity after t seconds is (48-3t)cm/s. If its distance from A after t seconds is s cm, find s in terms of t. Hence find the time that elapses before the particle is back again at A.

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Answer:


\sf s=\left(24t-\frac32t^2\right) \ cm

16 s

Explanation:

Given:

  • s = s
  • u = 0
  • v = 48 - 3t
  • a =
  • t = t

Substituting values into SUVAT formula to find s in terms of t:


\sf s=\frac12(u+v)t


\implies \sf s=\frac12(0+48-3t)t


\implies \sf s=\frac12(48-3t)t


\implies \sf s=24t-\frac32t^2

To find the time that elapses before the particle is back at A, set s to zero and solve for t:


\sf \implies 24t-\frac32t^2=0


\sf \implies t(24-\frac32t)=0


\sf So \ t=0 \ and \ 24-\frac32t=0


\sf \implies 24-\frac32t=0


\sf \implies \frac32t=24


\sf \implies t=16

Therefore, t = 0 and t = 16, so the time that elapses before the particle is back again at A is 16 s

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