Question

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.

Answer 1

Answer:

$\sf s=\left(24t-\dfrac32t^2\right) \ cm$

16 s

Step-by-step explanation:

Given:

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

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

      $\sf s=\dfrac12(u+v)t$

$\implies \sf s=\dfrac12(0+48-3t)t$

$\implies \sf s=\dfrac12(48-3t)t$

$\implies \sf s=24t-\dfrac32t^2$

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

$\sf \implies 24t-\dfrac32t^2=0$

$\sf \implies t(24-\dfrac32t)=0$

$\sf So \ t=0 \ and \ 24-\dfrac32t=0$

$\sf \implies 24-\dfrac32t=0$

$\sf \implies \dfrac32t=24$

$\sf \implies t=16$

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