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

`\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