Question

A test rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 49 m and acquired a velocity of 30 m/s How long did the burn phase last? A: 2.6 s B: 2.3 s C: 3.3 s D:3.6 s E: 2.9 s

Answer 1

The rocket starts at rest, so its initial velocity is 0. If the burn phase lasts
`t` seconds, then during this interval the rocket's velocity is

`v=at\implies a=\frac{30\,(\mathrm m)/(\mathrm s)}t`

and its position is

`x=\frac12at^2\implies a=(2(49\,\mathrm m))/(t^2)`

So we have

`\frac{30\,(\mathrm m)/(\mathrm s)}t=(2(49\,\mathrm m))/(t^2)\implies t=3.3\,\mathrm s`

and the answer is C.