The distance covered by the rocket in the first 5 seconds is 7.13 m
Explanation:
In order to solve the problem, we have to find the acceleration of the rocket. We can do it by using Newton's second law, which states that the net force on the rocket is equal to the product between its mass (m) and its acceleration (a):
[tex]F(t)=ma(t)[/tex]
Where here we have:
[tex]F(t)=16.8t[/tex] is the time-dependent force
m = 49.5 kg is the mass
Solving for a(t),
[tex]a(t)=\frac{F(t)}{m}=\frac{16.8t}{49.5}=0.34t[m/s^2][/tex]
This is the time-dependent acceleration of the rocket.
By integrating this expression, we find the velocity of the rocket at time t:
[tex]v(t)=\int a(t)dt=\int (0.34t)dt=\frac{0.34t^2}{2}=0.17t^2[/tex]
where we didn't add the constant term since the velocity at t = 0 is zero (the rocket starts from rest).
By integrating this expression, we then find an expression for the position of the rocket at time t:
[tex]x(t)=\int v(t) dt = \int (0.17t^2)dt=\frac{0.17t^3}{3}=0.057t^3[/tex]
Therefore, the distance covered by the rocket in the first t = 5.00 s is:
[tex]x(5)=0.057(5.0)^3=7.13 m[/tex]
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