A small 7.20-kg rocket burns fuel that exerts a time-varying upward force on the

Question

A small 7.20-kg rocket burns fuel that exerts a time-varying upward force on the rocket. This force obeys the equation F = A + Bt^2. Measurements show that at t = 0, the force is 133.0 N, and at the end of the first 1.50 s, it is 164.0 N. Find the constant A. Find the constants B. Find the net force on this rocket the instant after the fuel ignites. Find the acceleration of this rocket the instant after the fuel ignites. Find the net force on this rocket 3.30 s after fuel ignition. Find the acceleration of this rocket 3.30 s after fuel ignition. Suppose you were using this rocket in outer space, far from all gravity. What would its acceleration be 3.30 s after fuel ignition?

Explanation / Answer

Here ,

F = A + B*t^2

part A)

at t = 0

133 = A + B * 0

A = 133 N

the consant A is 133 N

B)

for the constant B

at t = 1.5 s

164 = 133 + 1.5^2 * B

B = 13.8 N/s^2

the constant B is 13.8 N/s^2

part C)

net force = F - m*g

net force = 133 + 13.8 * t^2 - 7.2 * 9.8

net force = 57.54 + 13.8 t^2

the net force is 57.54 + 13.8 t^2

at t = s

net force = 57.54 N

part D)

acceleration = net force/m

acceleration = 57.54/7.2

acceleration = 8 m/s^2

the acceleration of this rocket is 8 m/s^2

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