Question

A rocket initially at rest accelerates at a rate of 99.0 meters/second2. calculate the distance covered by the rocket if it attains a final velocity of 445 meters/second after 4.50 seconds.

Answer 1

Initial velocity(u) = 0m/s. Acceleration(a)= 99m/s^2. Final velocity(v)= 445m/s. Time taken by rocket to cover s distance = 4.5s. Using kinematic equation v^2 - u^2 = 2as, (445)^2 - 0^2 = 2x99xs. s = 445^2 / 2x99 = 1000.12 m. Hope it helps. 

Answer 2

Something must be wrong in the data you have, since this is basic using of linear motion's formulas.
vf=v0+at. Where vf= final velocity; v0= initial velocity, a=acceleration and t=time.
If the rocket is initially at rest, v0=0. Therefore vf=at. Plugging numbers in gives 445=99*4.5, However
445≠445.5.

Check it and then calculate the distance from x=a*t^2.