Question

A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 18 s, then the motor stops. The rocket altitude 22 s after launch is 5100 m. You can ignore any effects of air resistance.
What was the rocket's acceleration during the first 18 s?

Answer 1

http://bestdamntutoring.com/Rocket-Propulsion-Kinematics.html

This video might help you. You'd just need to switch the variables around.

Answer 2

d1 = 1/2 *a *(18)^2
Vf = a(18)

d2 = Vf*4 -1/2*g*(4)^2
d2 = a(18)*4 -1/2*g*(4)^2

d1 +d2 = 5100 =1/2 *a *(18)^2 + a(18)*4 -1/2*g*(4)^2

Answer 3

solve for 'a' and stuff

Answer 4

Let d denote the total displacement, d1 denote the distance covered under positive acceleration, d2 denote the distance covered after the motor stops, a1 denote the acceleration in the first 18s.
d = 5100m = s1 + s2
d1 = ?
d2 = ?
a1 = ?
a2 = gravity = 9.8 m/s^2 [down]

Time under positive acceleration squared = (18s)^2 = 324

Distance traveled under positive acceleration \[d=\frac{ a_{1}*t_{1} ^{2} }{ 2 }=\frac{324a_{1}}{2} =162a_{1}\]
Final velocity after acceleration = \[v _{f}= a_{1}*t_{1} + u = 18a_{1} + 0 = 18a_{1}\]
\[u = v _{f} = 18a_{1}\]
Acceleration under power:
\[d _{2}=u*t_{2} + \frac{ a_{2}*t_{2} }{ 2 }=18a_{1}*4 +\frac{a_{2}*4}{2}= 72a_{1} +\frac{-9.8*4}{2}=72a_{1}-19.6 \]
\[d=5100 = d _{1}+d_{2}=162a_{1} + 72a_{1}-19.6 = 234a_{1}-19.6\]
\[5100+19.6 = 234a_{1} -19.6 + 19\]
\[a_{1}=\frac{ 5119.6 }{ 234 } = 21.88\approx21.9\]
Therefore, the acceleration during the first 18s was approximately 21.9 m/s^2