Question

Question #05757

Answer 1

`6.8" ms"^-1`

Explanation:

It is a case of non-uniform acceleration. (Assuming that `t` is acceleration is time).
We know that
Acceleration `a=(dv(t))/dt=-0.9t`
`=>dv(t)=-0.9tcdot dt`

Integrating both sides we get
`=>v(t)=-0.9int tcdot dt`
`=>v(t)=-0.9(t^2/2+C)` .....(1)
where `C` is a constant of integration. It is given that the initial velocity at `t=0` is `23" ms"^-1`. Inserting in (1)
`v(0)=23=-0.9(0^2/2+C)`
`=>-0.9C=23`
`=>C=-23/0.9`
Expression for velocity becomes
`=>v(t)=-0.9(t^2/2-23/0.9)`
`=>v(t)=-0.45t^2+23` ......(2)

To find the velocity after `6s`. From (2) we get
`v(6)=-0.45xx6^2+23`
`v(6)=6.8" ms"^-1`