A rocket begins its third stage of launch at a velocity
of 2.28
102 m/s [fwd]. It undergoes a constant
acceleration of 6.25
101 m/s2, while travelling
1.86 km, all in the same direction. What is the rocket’s
velocity at the end of this motion?
To solve this problem, we can use the kinematic equation:
vf^2 = vi^2 + 2*a*d
Where:
vf = final velocity
vi = initial velocity
a = acceleration
d = distance
Given:
vi = 2.28 * 10^2 m/s
a = 6.25 * 10^1 m/s^2
d = 1.86 * 10^3 m
Plugging in the values:
vf^2 = (2.28 * 10^2)^2 + 2 * 6.25 * 10^1 * 1.86 * 10^3
vf^2 = 5,198.4 + 2 * 6.25 * 1.86 * 10^3
vf^2 = 5,198.4 + 23250
vf^2 = 27,448.4
vf = sqrt(27,448.4)
vf ≈ 165.8 m/s
Therefore, the rocket's velocity at the end of this motion is approximately 165.8 m/s [fwd].