Question

A rocket initially at rest acceleration at a rate of 99.0meter/second^2. Calculate the distance covered by the rocket if it attains a final velocity of 445 meters/seconds after 4.50 seconds.

Answer 1

Distance traveled by the rocket= 1002 m

Explanation:

initial velocity= Vi=0

final velocity= V=445 m/s

time=4.5 s

Acceleration=a=99 m/s²

using kinematic equation

h= Vi t + 1/2 at²

h= 0(4.5) + 1/2 (99) (4.5)²

h=1002 m

Thus the distance traveled by the rocket= 1002 m