The rocket's acceleration for the first 16 s is 27 m/s².
The rocket starts from rest and experiences uniform upward acceleration a for a time t₁ = 16 s. During this period it travels upwards a distance s₁.
Use the equation of motion
[tex]s=ut+\frac{1}{2} t^2[/tex]
Substitute s₁ for s, 0 m/s for u, 16 s for t =t₁, and write an equation for s₁ in terms of a.
[tex]s=ut+\frac{1}{2} t^2\\ s_1=(0 m/s)(16s)+\frac{1}{2} a(16s)^2\\ s_1=128a........(1)[/tex]
The engines are switched off after 16 s and for the next 4 s, the rocket travels upwards under the acceleration due to gravity g, which is directed down wards.
Write an expression of the velocity v of the rocket at the end of 16 s.
Use the equation of motion,
[tex]v=u+at[/tex]
Substitute 0 m/s for u and 16 s for t =t₁,
[tex]v=u+at\\ v=(0m/s)+a(16 s)\\ v=16a........(2)[/tex]
The rocket has a speed of v =16a at the beginning of its motion after its engines are switched off.
Determine the distance s₂ the rocket travels under the action of acceleration due to gravity g.
Use the equation of motion
[tex]s=ut+\frac{1}{2} t^2[/tex]
Substitute s₂ for s, 16 a for u, 4 s for t and -9.8 m/s² for g.
[tex]s=ut+\frac{1}{2} t^2\\ s_2=(16a)(4 s)-(9.8 m/s^2)(4 s)^2\\ s_2=(64a)-(78.4 m).......(3)[/tex]
The total distance s traveled by the rocket is given by,
[tex]s=s_1+s_2[/tex]
Add equations (1) and (3) and substitute 5100 m for s.
[tex]s=s_1+s_2\\ 5100m=128a+64a-(78.4 m)\\ 192 a = 5178.4\\ a= 26.97 m/s^2[/tex]
The acceleration of the rocket in the first 16 s is 27 m/s^2(2sf)
