Question

A partical moves in a straight line as `s=alpha(t-2)^(3)+beta(2t-3)^(4)`, where `alpha` and `beta` are constants. Find velocity and acceleration as a function of time.

Answer 1

`s=alpha(t-2)^(3)+beta(2t-3)^(4)`
`=alpha{(t-2)^(3)}+beta{(2t-3)^(4)}`
`v=(ds)/(dt)=alpha.3(t-2)^(2).(1-0)+beta.4(2t-3)^(3).(2.1-0)`
`=3alpha(t-2)^(2)+8beta(2t-3)^(3)`
`=3alpha{(t-2)^(2)}+8beta{(2t-3)^(3)}`
`a=(dv)/(dt)=3alpha.2(t-2).(1-0)+8beta.3(2t-3)^(2).(2.1-0)`
`=6alpha(t-2)+48beta(2t-3)^(2)`
Recall: First differential the outer layer, then the inner layer and then multiply the two differentials.