Question
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?
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?
Answers
GPT 3.5
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].
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].